Unit 6 - Problem 8 ==> Charge Detective

Given:

• $$ \mbox{Bottom left charge} = q $$
• $$ \mbox{Call the top charge} = q_1 $$
• $$ \mbox{Call the bottom right charge} = q_2 $$
• $$ 23.41^\circ = \mbox{combined force of } $$
• $$ 90^\circ \; \mbox{is combined force on q_2} $$
• $$ \mbox{The three charges make an equilateral triangle of side} = L $$

The Question:

• $$ \mbox{Using the q charge as the base, what are the relative values of the other two charges?}$$

Rationale: Approach using approximation

• $$ q_1 \; \mbox{and} q_2 \; \mbox{have the same polarity since their forces are pushing them apart.}$$
• $$ q \; \mbox{has opposite polarity to } \; q_1 \mbox{and}\; q_2 \; \mbox{since the 23.41 degree force is towards} \; q_1 \mbox{and}\; q_2 $$
• $$ \mbox {If} \;  q_1 = q_2 \;\mbox{then the angle at q would be 30 degrees} $$
• $$ \frac{q_2}{q_1} \varpropto \Delta\mbox{from 30 degrees} $$
• $$ \frac{30^\circ}{60^\circ} = \frac{1}{2} $$
• $$ \frac{23.41^\circ}{60^\circ} = 0.39 $$
• $$ \Delta = 0.50 - 0.39 = 0.11 $$
• $$ q_2 = 1.11 * q_1 $$
• $$ \mbox{Extending a line from} \; q_1 \; \mbox{thru} \; q_2 $$ 
• $$ \mbox{this force is the hypotunese of a right triangle with angle of }\; 30^\circ $$
• $$ \sin{30}^\circ = \frac{\frac{K * Q_2 * Q_3}{r^2}}{\frac{K * q_1 * q_2}{r^2}}  $$ 
• $$  \sin{30}^\circ = \frac{q_3}{q_1} $$

Calculate: Using approximation approach

• $$ q_1 = \frac{q_3}{\sin{30}^\circ} = 2 * q_3 $$
• $$ q_2 = 1.11 * q_1 = 1.11 * \frac{q_3}{\sin{30}^\circ} =  2.22 * q_3 $$

Rationale: Approach only based on electric charges

 

Definitions

• $$ \alpha = \mbox{sixty degree angle due to the equilateral arrangement of the three charges} $$
• $$ \beta = \mbox{23.41 degree angle of attraction of q to the other two charges} $$
• $$ \theta = \mbox{30 degree angle pulling on the bottom right charge} $$

Process the forces on charge q

• $$ F_{1-3} = \frac{K * q_1 * q_3}{L^2} $$
• $$ \cos\alpha = \frac{F_{1-3-X}}{F_{1-3}} $$
• $$ F_{1-3-X} = F_{1-3} \cos\alpha $$
• $$ F_{1-3-X} = \cos\alpha *  \frac{K * q_1 * q_3}{L^2}$$
• $$ \sin\alpha = \frac{F_{1-3-Y}}{F_{1-3}} $$
• $$ F_{1-3-Y} = F_{1-3} \sin\alpha $$ 
• $$ F_{1-3-Y} = \sin\alpha * \frac{K * q_1 * q_3}{L^2} $$
• $$ F_{2-3} = F_{2-3-X} = \frac{K * q_2 * q_3}{L^2} $$
• $$ F_{2-3-Y} = 0 $$
• $$ \sin\beta = \frac{F_{total-1-2-Y} }{F_{total-1-2} } $$
• $$ \cos\beta = \frac{F_{total-1-2-X} }{F_{total-1-2} } $$
• $$ F_{total-X} = F_{1-3-X} + F_{2-3-X}$$
• $$ F_{total-X} = \left( \cos\alpha *  \frac{K * q_1 * q_3}{L^2} \right)  + \frac{K * q_2 * q_3}{L^2} $$ 
• $$ F_{total-X} = \left( \frac{K * q_3}{L^2} \right) * ((q_1 * \cos\alpha) + q_2) $$
• $$ F_{total-Y} = F_{1-3-Y} + F_{2-3-Y} =  F_{1-3-Y} $$
• $$ F_{total-Y} = \sin\alpha * \frac{K * q_1 * q_3}{L^2} $$
• $$ \tan\beta = \frac{F_{total-y}}{F_{total-X}} $$
• $$ \tan\beta = \frac{\sin\alpha * \frac{K * q_1 * q_3}{L^2}}{\left( \frac{K * q_3}{L^2} \right) * ((q_1 * \cos\alpha) + q_2)}$$
• $$ \tan\beta = \frac{\sin\alpha * q_1}{ (q_1 * \cos\alpha) + q_2)}$$
• $$ \tan\beta * \cos\alpha * q_1 + \tan\beta * q_2 = q_1 * \sin\alpha $$
• $$\tan\beta * q_2 = (q_1 * \sin\alpha) - (q_1 * \tan\beta * \cos\alpha)) $$
• $$\tan\beta * q_2 = q_1 * (\sin\alpha - (\tan\beta * \cos\alpha)) $$
• $$ q_2 = \frac{q_1 * (\sin\alpha - ( \tan\beta * \cos\alpha))}{\tan\beta}$$

Process the two opposing charges

• $$ \tan\theta = \frac{F_{2-3}}{F_{1-2}} $$
• $$ \tan\theta = \frac{\frac{K * q_2 * q_3}{L^2}}{\frac{K * q_1 * q_2}{L^2}} $$
•  $$ \tan\theta = \frac{q_3}{q_1} $$
• $$ q_1 = \frac{q_3}{\tan\theta} $$

Calculate: Approach only based on electric charges

• $$  \mbox{The angle data:} $$
• $$ \sin\alpha = \sin{60^\circ} = 0.866025 $$
• $$ \cos\alpha = \cos{60^\circ} = 0.500 $$
• $$ \tan\beta = \tan{23.41^\circ} = 0.432946 $$
• $$ \tan\theta = \tan{30^\circ} = 0.57735 $$ 
• $$ q_2 = \frac{q_1 * (\sin\alpha - ( \tan\beta * \cos\alpha))}{\tan\beta}$$
• $$ q_2 = q_1 * \frac{(.866025 - (.432946 * .5))}{.432946} $$
• $$ q_2  = q_1 * \frac{.649552}{.432946} = q_1 * 1.500 $$
• $$ q_1 = \frac{q_3}{\tan\theta} $$
• $$ q_1 = q_3 * \frac{1}{\tan\theta} = 1.732q_3 $$
• $$ q_2 = 1.5  q_1 = 1.5 * 1.732 * q_3 = q_3 * 2.598 = 2.6q_3 $$

 

Comparing Approaches

Charge	Approximation	Electric Charges	% Diff
q top	Q1 = 2*Q3	Q1 = 1.732*Q3	15.5 %
q right	Q2 = 2.22*Q3	Q2 = 2.6*Q3	14.6 %

Unit 6 - Problem 7 ==> Fields and Angles

Given:

• $$ q_1, q_2, \theta_1, \theta_2 $$

Rationale:

• $$ \tan\theta_1 = \frac{E * q_1}{F_{both}}  $$
• $$ F_{both} = \frac{E * q_1}{\tan\theta_1}  $$
• $$ \tan\theta_2 = \frac{-E * q_2}{-F_{both}}  $$ 
• $$ F_{both} = \frac{E * q_2}{\tan\theta_2}  $$
• $$ F_{both} = F_{both} $$
• $$ \frac{E * q_1}{\tan\theta_1}= \frac{E * q_2}{\tan\theta_2}  $$
• $$ \frac{q_1}{q_2} = \frac{\tan\theta_1}{\tan\theta_2} $$

Calculate

• $$ \frac{q_1}{q_2} = \frac{\tan{76}^\circ}{\tan{24}^\circ} $$
• $$ \frac{q_1}{q_2} = \frac{4.01078}{.445229}  = 9.0 $$

Unit 6 - Problem 6 ==> Battle of the Forces

Given:

• $$ K_{electrostatic-constant} = 8.99 * 10^9 $$
• $$ G_{gravity-constant} = 6.67 * 10^-11 $$
• $$\mbox{Two bodies with same mass and charge of 1 coulomb each.}$$

Question:

• $$\mbox{At what mass are the force of gravity and the electrostatic force equal?}$$

Rationale:

• $$F_{gravity} = F{electrostatic-charge} $$
• $$ \frac{G * m^2}{r^2} = \frac{K * Q^2}{r^2} $$
• $$ m^2 = \frac{K * Q^2}{G} $$
• $$ m = \sqrt\frac{K * Q^2}{G} $$

Calculate

• $$ m = \sqrt\frac{(8.99 * 10^9)}{(6.67 * 10^-11)} $$
• $$ m = \sqrt{(1.347826 * 10^{20})} $$
• $$ m = 1.16 * 10^{10} $$

Comparison:

• $$ \mbox{mass of the earth} = 5.972 * 10^{24}\;\mbox{kg} $$
• $$ \frac{ 1.16 * 10^{10}}{5.972 * 10^{24}} = 0.19*10^{-14} $$
• $$ \mbox{Much smaller than the earth} $$
• $$ \mbox{mass of the earth's moon} = 0.07349 * 10^{24} $$
• $$ \frac{ 1.16 * 10^{10}}{0.7349 * 10^{24}} =15.7844604 * 10{-14} $$
• $$ \mbox{Much smaller than earth's moon} $$
• $$ \mbox{mass of a US Navy nuclear-powered aircraft carrier} = 196 * 10^6 \;\mbox{lbs} $$
• $$ \mbox{number of carriers} = \frac{1.16 * 10^{10}}{196 * 10^6} = 59.18367 $$
• $$ \mbox{That's more carrier's than the US Navy has ever had.} $$

Unit 6 - Problem 5 ==> Electropendulum

Given:

• $$ L = 0.75 \;\mbox{meters}  $$
• $$ m = 0.02 \;\mbox{kg}  $$
• $$ E = 10 \;\mbox{n/s} $$
• $$ q = 0.01 \;\mbox{coulombs} $$

Rationale:

• $$ T = 2 \pi \sqrt\frac{L}{a} $$
• $$ F = m * a $$
• $$ a = \frac{F}{m} $$ 
• $$ T = 2 \pi \sqrt\frac{L}{\frac{F}{m} } = 2 \pi \sqrt\frac{(m * L)}{F} $$
• $$ F_e = E * q $$
• $$ T =  2 \pi \sqrt\frac{(m * L)}{(E * q)} $$

Calculate

• $$ T =  2 \pi \sqrt\frac{(0.02 * 0.75)}{(10.0 * 0.01)} $$ 
• $$ T =  2 \pi \sqrt\frac{0.015}{ 0.1} $$
• $$ T =  2 \pi \sqrt{0.15} $$ 
• $$ T =  2 \pi * 0.387298 $$ 
• $$ T = 2.43 \;\mbox{seconds} $$

Unit 6 - Problem 4 ==> Cathode Ray Tube

Given:

• $$ d = 0.5 \;\mbox{meters} $$
• $$  V_0 = 1 * 10^7 \;\mbox{meters/second}   $$
• $$ q = 1.602 * 10^{-19} \;\mbox{coulombs} $$
• $$ m_e = 9.11 * 10^{-31} \;\mbox{kg} $$

Question:

• $$\mbox{What electrical field is required to cause a displacement} = \Delta{y} \;\mbox{?} $$

Overview of Approach:

1. Determine horizontal time
2. Determine the vertical acceleration
3. Determine electrical field necessary for vertical displacement
4. Finish by substituting for the time to transit

Rationale:

Determine horizontal time

• $$ d = V_0 * t_x $$
• $$ t _x = \frac{d}{V_0} $$

Determine the vertical acceleration

• $$ F_{movement}  = F_e $$
• $$ F_{movement} = m * a $$
• $$ F_e = E * q $$
• $$ F_{movement} = F_e $$
• $$ E * q = m * a $$
• $$ a_y = \frac{(E * q)}{m} $$

Determine electrical field necessary for vertical displacement

• $$ t_x = t_y $$


• $$ \Delta{y} = \frac{1}{2}  * a_y * t_y^2$$
• $$ \frac{(2 * \Delta{y})}{t_x^2} = a_y = \frac{(E * q)}{m} $$
• $$ E = \frac{(m * 2 * \Delta{y})}{(q * t_x^2)}$$

Finish by substituting for the time to transit

• $$ t_x^2 = \frac{d^2}{V_0^2}  $$
• $$ E = \frac{(V_0^2 * m * 2 * \Delta{y})}{(q * d^2)} $$

Calculate

• $$ E = \frac{(10^{14} * 9.11 * 10^{-31} * 2 * 0.1)}{(1.602 * 10^{-19} * 0.25)}  $$
• $$ E = \frac{(10^{-17} * 1.822)}{(10^{-19} * 0.4005)}  $$
• $$ E = 10^2 * 4.54931 = 454.931 = 455 $$

Unit 6 - Problem 3 ==> Parallel Plates Potential Energy

Given:

• $$ \mbox{d, x, q, m, E} $$

Question:

• $$\mbox{What is the electrical potential U at x?} $$

Rationale:

• $$ U_e = F_e * x $$
• $$ F_e = E * q $$

Calculate

• $$ U_e = E * q * x  $$
• $$ U_{e-max} = E * q * d  $$

References:

• Wikipedia

Unit 6 - Problem 2 ==> Parallel Plates

Given:

• $$ E_{field} = 1000 $$
• $$ d = 1 \; \mbox{cm} $$
• $$ m_{proton} = 1.673 * 10^{-27} $$
• $$ q = 1.6024 * 10^{-19} $$ 
• $$ \mbox{Initial velocity of charge}\; = 0 $$

Question:

• $$\mbox{How long, in nanoseconds, does it take the proton to cross between the two plates?} $$

Overview of Approach:

1. Convert centimeters to meters for final calculations
2. Determine force on charge due to electrical field
3. Determine acceleration on charge due to force on charge
4. Determine time to transit using the acceleration

Rationale:

Convert centimeters to meters

• $$ 1 \; \mbox{cm} = 1 * 10^{-2} \; \mbox{meters} $$

Determine force on charge due to electrical field

• $$ E = \frac{F_e}{q} $$
• $$ F_e = E * q $$

Determine acceleration on charge due to force on charge

• $$ F = m * a $$
• $$ a = \frac{F}{m} $$
• $$ a_p = \frac{F_e}{m} = \frac{E * q}{m} $$

Determine time to transit using the acceleration

• $$ \Delta{x} = (V_0 * t)  + (\frac{1}{2} * a * t^2) $$
• $$ t^2 = \frac{2\Delta{x}}{a} $$
• $$ t^2 = \frac{(2 * d)}{ \frac{E * q}{m}}$$
• $$ t^2 = \frac{(2 * d * m)}{(E * q)} $$
• $$ t = \sqrt \frac{(2 * d * m)}{(E * q)} $$

Calculate

• $$ t = \sqrt \frac{(2 * (10^{-2}) * ( 1.673 * 10^{-27}))}{(1000 * (1.6024 * 10^{-19} ))} $$
• $$ t = \sqrt\frac{(3.346 * 10^{-29})}{(1.6024 * 10^{-16})}$$
• $$ t = \sqrt{(2.088 * 10{-13})} $$
• $$ t = .4569 * 10^{-6} =  456.9 * 10^{-9}$$

Unit 6 - Problem 1 ==> Frantic Fields

Given:

• $$ $$

Rationale:

• $$ $$

Calculate

• $$ $$

Unit 5 - Problem 8 ==> Springs In Series

Given:

• Two springs in series with spring constants k-1 and k-2.
• mass of m attached to spring 2.

Rationale:

• The k of the combined springs must be less than the k of either spring
• $$ F_g = m g    $$
• $$ F_{spring_2}  = m g $$
• $$ \Delta{x_{total}} = \Delta{x_2} + \Delta{x_1} $$
• $$\mbox{In general:  } \Delta{x} = \frac{F}{k} $$
• $$ \frac{m g}{k_{total}} = \frac{m g}{k_2} + \frac{m g}{k_1} $$
• $$ \frac{1}{k_{total}} =  \frac{1}{k_2} + \frac{1}{k_1} $$
• $$ \frac{1}{k_{total}} =  \frac{(k_2 + k_1)}{(k_1 * k_2)}  $$
• $$ k_{total} = \frac{(k_1 * k_2)}{(k_2 + k_1)} $$

Testcase: k-1 = k-2

• $$ k_1 = k_2 $$
• $$ k_{total}  = \frac{k_1^2}{2 k_1} $$
• $$ k_{total} = \frac{k_1}{2} $$

Testcase: k-1 <<  k2

• $$ k_1 = 100 $$
• $$ k_2 =  100,000 $$
• $$ k_{total} = \frac{(k_1 * k_2)}{(k_2 + k_1)} $$
•  $$ k_{total} = \frac{(100 * 100000)}{(100000 + 100)} $$
•  $$ k_{total} = 99.9 $$

Other References

• From Wolfram

Unit 5 - Problem 7 ==> Swing Set

Given:

• $$ l_{swing} = 4 $$
• $$ h_{swing} = 1 $$
• $$ m_{person} = 60 $$
• $$ {PE}_{max}  \mbox{is at} 45^\circ   \mbox{back} $$
• $$ \mbox{Jump off is at } 30^\circ  \mbox{forward} $$

Overview of Approach:

1. Consider energy available
2. Convert PE to KE and calculating x and y vectors
3. Calculate total time in air
4. Calculate horizontal distance from take-off

Rationale:

• Considering energy available
• $$ h_{PE_{max}} = l_{swing} \cos45^\circ $$
• $$ PE_{max} = m * g *  h_{PE_{max}} = m * g * l_{swing} \cos45^\circ $$
• $$ h_{jump} = ( l_{swing} + h_{swing} ) - ( l_{swing} \cos30^\circ ) $$
• $$ PE_{jump} = m * g * h_{jump} = m * g * ( ( l_{swing} + h_{swing} ) - ( l_{swing} \cos30^\circ )) $$
• $$ PE_{take-off} = PE_{max} - PE_{jump}$$
• Converting PE to KE and calculating x and y vectors
• $$ KE_{take-off} = PE_{take-off} $$
• $$ \frac{1}{2} m V_0^2 = PE_{take-off} $$
• $$ V_0^2 = \frac{2 PE_{take-off}}{m} $$
• $$ V_0 = \sqrt \frac{2 PE_{take-off}}{m} $$
• $$ V_{0-x} = V_0 \cos30^\circ $$
• $$ V_{0-y} = V_0 \sin30^\circ $$
• Calculate total time in air
• $$ t_{up} = -\frac{(V - V_{0-y})}{a} $$
• $$ t_{up} = -\frac{-V_{0-y}}{g} = \frac{V_{0-y}}{g}$$
• $$ \Delta{y_{up}} = t_{up} \frac{V + V_{0-y}}{2} = t_{up} \frac{V_{0-y}}{2} $$
• $$ \Delta{y_{down}} =  \Delta{y_{up}} + h_{jump} $$
• $$ \Delta{y_{down}} = V t + \frac{1}{2} a t_{down}^2 = \frac{1}{2} g t_{down}^2 $$
• $$ t_{down}^2 = \frac{2 \Delta{y_{down}}}{g}   $$
• $$ t_{down} = \sqrt \frac{2 \Delta{y_{down}}}{g} $$
• $$ t_{total} = t_{up} + t_{down} $$
• Calculate horizontal distance from take-off
• $$ \Delta{x} = V_{0-x} * t_{total} $$

Calculations:

• $$ \cos45^\circ = .707107 $$
• $$ \cos30^\circ = .866025 $$
• $$ \sin30^\circ = .500000 $$
• $$ \Delta{x} = 2.35  \mbox{meters} $$

Unit 5 - Problem 6 ==> Block Clock

Given:

• $$ \mbox{mass}  = 0.5  \mbox{kg} $$
• $$ k = 100 $$
• $$ \mbox{distance between springs} = 3  \mbox{meters} $$
• Switch  at midpoint and need to hit every second

Rationale:

• Ignore the width of the block
• Sequence: 1.5 m to spring, bounce on spring, then 1.5 m back to switch -- then again on opposite side
• $$ V_0 t = 2 * 1.5  \mbox{meters}  $$
• $$ t = \frac{3}{V_0}  $$
• Uses 0.5 of spring period
• $$ \frac{1}{2} T_{spring} = \frac{1}{2} (2 \pi) \sqrt\frac{m}{k}  $$
• $$ t +\frac{T}{2} = 1  \mbox{second} $$
• $$ \frac{3}{V_0} + \pi \sqrt\frac{m}{k} = 1  $$
• $$ \frac{3}{V_0}  = 1 - \pi \sqrt\frac{m}{k}  $$
• $$ V_0 = \frac{3}{( 1 - \pi \sqrt\frac{m}{k})} $$

Calculate:

• $$ V_0 =   \frac{3}{( 1 - \pi \sqrt\frac{0.5}{100})} $$
• $$ V_0 =   \frac{3}{( 1 - \pi * (.07071067811))} $$
• $$ V_0 =   \frac{3}{( 1 - 0.222 )} $$ 
•  $$ V_0 =   \frac{3}{.778} $$
• $$ V_0 = 3.8560  \mbox{meters/second} $$

Unit 5 - Problem 5 ==> Falling Thru The Earth

Given:

• $$ F_e = g m \frac{r}{r_e}  $$
• $$ g = 10 $$
• $$ r_e = 6,400  \mbox{km} = 6,400,000  \mbox{meters} $$

Rationale:

• $$ F_{restoring} = k x = r \frac{g m}{r_e} $$
• $$ k = \frac{g m}{r_e} $$
• $$ \frac{1}{k} = \frac{r_e}{g m} $$
• $$ x = r $$ 
• $$ T_{period} = 2 \pi \sqrt{(m \frac{1}{k})} $$
• $$ T_{period} = 2 \pi \sqrt{(m \frac{r_e}{g m})} $$
• $$ T_{period} = 2 \pi \sqrt\frac{r_e}{g} $$

Calculate:

• $$ T_{period} = 2 \pi \sqrt\frac{r_e}{g} $$
• $$ T_{period} = \frac{44}{7} \sqrt\frac{6,400,000}{10} $$
• $$ T_{period} = 6.2857143 \sqrt{640,000} $$
• $$ T_{period} = 6.2857143 * 800 = 5,028.57  \mbox{seconds}$$
• $$ T_{period} =83.8  \mbox{minutes} $$

Unit 5 - Problem 4 ==> Shock Absorber Damping

Given:

• 20.6% of energy lost per cycle of shock absorber
• $$ E_n = (1 - .206)^n E_0  = (0.794)^n E_0 $$
• $$ E \varpropto A^2 $$

Rationale:

• $$ (.5 A)^2 \varpropto E
• $$ (0.794)^n = .5^2 = .25 $$

Calculate:

• Powers of 0.794
• $$ 0.794 ^2 = 0.630436 $$
• $$ 0.794 ^3 = 0.500566$$
• $$ 0.794 ^4 = 0.397449 $$
• $$ 0.794 ^5 = 0.315574 $$
• $$ 0.794 ^6 = 0.250566 $$
• n = 6

Unit 5 - Problem 3 ==> Shock Absorber Period

Given:

• $$ \mbox{mass of the car} = 2,000  \mbox{kilograms} $$
• $$ \mbox{four shock absorbers with four wheels}$$
• $$ \mbox{each shock absorber has }  k = 100,000 $$
• $$ \mbox{gravity}  = 10 $$ 

Rationale:

• $$ T_{period-shock-absorber} = 2 \pi \sqrt\frac{m_{on-shock-absorber}}{k} $$
• $$ m_{on-shock-absorber} = \frac{m_{car}}{4} $$
• $$ T_{period-shock-absorber} = 2 \pi \sqrt\frac{\frac{m_{car}}{4}}{k} = 2 \pi \sqrt\frac{m_{car}}{4 k} $$

Calculate:

• $$ T_{period-shock-absorber} = 2 \pi \sqrt\frac{2,000}{4 * 100,000} $$
• $$ T_{period-shock-absorber} = \frac{44}{7} \sqrt{.005} = 6.2857 * .07071 = 0.44 $$

Unit 5 - Problem 2 ==> Shock Absorbers

Given:

• $$ \mbox{mass of the car} = 2,000  \mbox{kilograms} $$
• $$ \mbox{four shock absorbers with four wheels}$$
• $$ \mbox{each shock absorber has }  k = 100,000 $$
• $$ \mbox{gravity}  = 10 $$

Rationale:

• $$ F_{car} = m_{car} g $$
• $$ F_{wheel} = \frac{F_{car}}{4} $$
• $$ F_{shock-absorber} = k \Delta{x} $$
• $$ F_{wheel} =  F_{shock-absorber}$$
• $$ \frac{F_{car}}{4}= k \Delta{x} $$
• $$ \Delta{x} = \frac{m_{car} g}{4 k} $$

Calculate:

• $$ \Delta{x} = \frac{2000 * 10}{4 * 100,000} $$
• $$ \Delta{x} = 0.05  \mbox{meters} = 5  \mbox{cm} $$

Unit 5 - Problem 1 ==> Keeping Time

Given:

• $$l_e = \mbox{length of pendulum on earth}  $$
• $$m = \mbox{mass of pendulum}$$
• $$g_e = \mbox{acceleration of gravity on the earth}  $$
• $$g_m = \mbox{acceleration of gravity on the moon}  $$

Rationale:

• $$T_{period-earth} = 2 \pi \sqrt\frac{l_e}{g_e}$$
• $$T_{period-moon} = 2 \pi \sqrt\frac{l_m}{g_m}$$
• $$T_{period-earth} = T_{period-moon} $$
• $$ 2 \pi \sqrt\frac{l_e}{g_e} = 2 \pi \sqrt\frac{l_m}{g_m}$$
• $$ \frac{l_e}{g_e} = \frac{l_m}{g_m}$$
• $$ l_m = l_e \frac{g_m}{g_e} $$

Calculate: Length of pendulum on moon

• $$ l_m = l_e \frac{1.6}{9.8} = 0.163265  l_e = 0.16  l_e $$

Simple Harmonic Equations

For springs	For pendulum	comments
$$F_{spring-restoring} = k \Delta{x} $$	$$F_{pendulum-restoring} = -m g\sin\theta $$
$$T_{period-of-oscillation} = 2\pi\sqrt\frac{m}{k} $$	$$T_{period-of-oscillation} = m g$$
$$k = m * \left(\frac{2\pi}{T}\right)^2 $$
$$m = k * \left(\frac{T}{2\pi}\right)^2 $$
$$x = A \cos (\omega t + \phi)$$		$$\omega \mbox{ is angular velocity in radians} $$ $$\phi \mbox{ initial angular displacement in radians} $$ $$\omega t \mbox{ displacement angle}$$
$$T = \frac{2 \pi}{\omega}$$
	$$\omega = \sqrt \frac{g}{l}$$
	$$T = 2 \pi \sqrt \frac{l}{g} $$

Hooke's Law & Simple Harmonic Mechanical Oscillation

Source

MIT Professor Lewin

Springs

Hooke's Law

• If we displace the spring a distance of x, then $$F_{spring-restoring} = -k x $$
• This is also known as Hooke's Law.

Measuring the spring constant k

• $$k = \frac{\Delta F}{\Delta X} $$

• $$T_{period-of-oscillation} = 2\pi\sqrt\frac{m}{k} $$
• Knowing the mass and the period T, $$ k = m * \left(\frac{2\pi}{T}\right)^2 $$
• Also knowing period T and spring constant k, $$ m = k * \left(\frac{T}{2\pi}\right)^2 $$
• Notice these measurements do not depend on gravity and can be done in  the weightlessness of space.
• $$ m a =  -k x $$
• $$ m \ddot{x} = -k x$$
• $$ m \ddot{x} + k x = 0$$
• $$ \ddot{x} + x \frac{k}{m}  = 0$$
• Our test function for x is:
• $$ x = A \cos (\omega t + \phi)$$
• A is the amplitude of the oscillation
• omega is the angular frequency in radians/sec
• psi is the phase angle in radians 
• If the time t is advanced to T, then omega * T is 2 pi = 360 degrees = one full cycle or period
• $$ T = \frac{2 \pi}{\omega} $$
• $$ f = \frac{1}{T} ~ \mbox{hertz}$$
• $$ \dot{x} = - A \omega \sin (\omega t + \phi)$$
• Derivative is with regard t
• Because the derivative of the cos is the -sin 
• And because omega is multiplied time t
• $$ \ddot{x} = - A \omega^2  \cos (\omega t + \phi)$$
• The derivative of the sin is the cos 
• $$ \ddot{x} = - A \omega^2  \cos (\omega t + \phi)$$
• Already started with $$ x = A \cos (\omega t + \phi) $$
• $$ \ddot{x} = - \omega^2  x$$
• We already have: $$ \ddot{x} + x \frac{k}{m}  = 0$$
• $$ - \omega^2  x + x \frac{k}{m}  = 0$$
• $$ \omega^2 =  \frac{k}{m}  = 0$$
• $$\omega = \sqrt \frac{k}{m} $$
• $$T = 2 \pi \sqrt\frac{m}{k} $$

Pendulum

• l = length of the string
• theta = the angle the pendulum is displaced from the vertical

• In the x direction: $$m a = m \ddot{x} = - T_\theta \sin\theta = - T_\theta \frac{x}{l} $$
• In  the y direction: $$ m \ddot{y} = T_\theta \cos\theta - m g $$
• These two equations are unsolvable as they are. But now we will make some approximations.
• We will use "Small Angle Approximations"
• Theta must remain a small angle. So theta must be << 1.
• Cosine of theta must be very close to 1; and sine theta will be very close to 0.
• 5 degrees = 0.996
• 10 degrees = 0.985
• The y acceleration is approximately zero
• The y excursion at 5 degrees is about 4% of x direction
• The y excursion at 10 degrees is about 9 % of x direction
• Revisiting our previous unsolvable equations
• $$ m \ddot{y} = T_\theta \cos\theta - m g $$
• Since cosine of theta, $$ 0 = T -mg $$
• $$ T = m g $$
• $$ m \ddot{x} = - T_\theta \sin\theta = - T_\theta \frac{x}{l} $$
• $$ m \ddot{x} + m g \frac{x}{l} = 0 $$
• $$ \ddot{x} + \frac{g}{l} x = 0 $$
• Notice the similarity of the spring equation and the pendulum equation
•  $$ \ddot{x} + \frac{k}{m} x  = 0  \mbox{for a spring}$$
•  $$ \ddot{x} + \frac{g}{l} x = 0  \mbox{for a pendulum}$$
• Since we have solved  the equation for the spring, the equations by extension for the pendulum are:
• $$ x = A \cos (\omega t + \phi)$$ 
• $$ omega = \sqrt \frac{g}{l} $$
• $$ T = 2 \pi \sqrt \frac{l}{g} $$

Unit 3 - Lecture 16 ==> Mass & acceleration relationship

Using the data given

• Decomposing  $$ \Delta{x} = V_0 t + \frac{1}{2} a t^2 $$ :
• Since the initial velocity is zero, $$ V_0 t = 0 $$
• Thus we have $$ \Delta{x} = \frac{1}{2} a t^2 $$
• Solving for a, $$a = \frac{2\Delta{x}}{t^2} $$
• Calculate 4 cases for acceleration using force of 10 Newtons

  Mass	Time	Acceleration	Acceleration Rounded
  10 kg	4.47	1.0009559129	1.00
  20 kg	6.32	0.500721038295	0.50
  30 kg	7.75	0.332986472425	0.33
  40 kg	8.94	0.250238978224	0.25

• $$\mbox{Acceleration} \varpropto \mbox{Mass}$$
• So find a relationship where the constant c is a constant
• Try $$ c =  a * m $$
• Calculate 4 cases for the constant c

Mass	Acceleration	Constant c
10	1.00	10
20	0.50	10
30	0.33	10
40	0.25	10

• And no other combination results in a constant constant
• Thus as a goes up or down, m must go down or up the same amount for F to  remain a fixed value.
• Thus F = c * m, and we have just calculated the force of gravity. 

Conclusions

• By observation, acceleration is directly proportional to the force applied.
• And as mass increases, acceleration proportionality decreases.
• $$ a \varpropto \frac{1}{m} $$
• And since $$ F = \mbox{constant} * m $$
• $$ c = 10 $$
• $$ F = 10 * m $$

Calculating the Coefficient of Friction using Inclined Plane

Given:

• The normal force is always perpendicular to the plane supporting the mass
• The y axis is perpendicular to the plane supporting the mass
• The x axis is perpendicular to the y axis
• mu is the dimensionless coefficient of friction
• The mu before the object moves is a static mu
• The mu after the object moves is a kinetic mu
• The static mu is always greater than the kinetic mu
• Force due to gravity toward the center of the earth is mg

Definition: General friction forces

• $$ F_{friction} = \mu {F_{normal}} $$
• $$ F_{gravity} = m g $$
• $$ F_{friction} = \mu m g$$

Definition: Forces acting on mass on inclined plane

• $$ \alpha_{angle-of-incline}$$
• $$ F_{normal} = F_y = mg\cos\alpha $$
• $$ F_{down-surface} = F_{friction}$$
• $$ F_{down-surface} = F_x = mg \sin\alpha $$
• $$ F_{friction} = \mu F_y = \mu mg \cos\alpha $$
• $$ mg\sin\alpha = \mu mg \cos\alpha $$
• $$ \mu = \frac{\sin\alpha}{\cos\alpha} $$
• $$ \sin = \frac{opposite}{hypotunese} $$
• $$ \cos = \frac{adjacent}{hypotunese} $$
• $$ \frac{\sin}{\cos} = \frac{\frac{opposite}{hypotunese}}{\frac{adjacent}{hypotunese}} $$
• $$ \frac{\sin}{\cos} = \frac{opposite}{hypotunese} \frac{hypotunese}{adjacent} $$
• $$ \frac{\sin}{\cos} = \frac{opposite}{adjacent} = \tan $$
• $$ \mu = tan\alpha $$
• So we now can use an adjustable inclined plane, raise the angle alpha until the object just breaks loose, measure the angle alpha, and set the static coefficient of friction equal to the tangent of angle alpha. Notice the static mu does not depend on:
• The mass of the object
• The surface area the object is in contact

Unit 4 - Problem 11 ==> Escape Velocity

Definition and Givens

• Escape velocity is the speed at which the kinetic energy plus the gravitational potential energy of an object is zero.
• It is the speed needed to "break free" from a gravitational field without further propulsion.
• According to Wikipedia, escape velocity equals the square root of 2 * G * m divided by r, where:
• G is the gravitational constant of 6.674 * 10^-11
• m is the mass of the body we are escaping from
• Mass of the earth is M=5.9736×10²⁴ kg
• r the distance from the center of gravity
• Radius of the earth is 6400 kilometers or 6.4 * 10^6 meters
• $$ V_{escape} = \sqrt{\frac{(2 G m)}{r}} $$
• On the surface of the Earth, the escape velocity is about 11.2 kilometers per second (~6.96 mi/s), which is approximately 34 times the speed of sound (Mach 34) and several times the muzzle velocity of a rifle bullet (up to 1.7 km/s).

Deriving the escape velocity formula

• Conservation of energy provides us that the escape energies total zero.
• $$ E_{initial-potential} + E_{initial-kinetic} = E_{escape-potential} + E_{escape-kinetic-energy} $$
• $$ E_{initial-potential} = -\frac{G M m}{r} $$
• $$ E_{initial-kinetic} = \frac{1}{2}m V^2  $$
• $$  -\frac{G M m}{r} + \frac{1}{2}m V_{escape}^2 = 0 + 0 $$
• $$  \frac{1}{2}m V^2 = \frac{G M m}{r}$$
• $$ V_escape^2 = 2\frac{G M}{r}$$
• $$ V_{escape} = \sqrt{\frac{(2 G M)}{r}} $$ 

Calculate the escape velocity value

• $$ V_{escape} = \sqrt{\frac{(2 G M)}{r}} $$
• $$ V_{escape} = \sqrt{\frac{(2* (6.674 * 10^-11) ( 5.97 * 10^{24}))}{6.4 * 10^6}} $$
• $$ V_{escape} = \sqrt{12.45118125 * 10^7} $$
• $$ V_{escape} = 11158.486120437665\frac{m}{s} = 24,960.8\;{mph}$$

MIT Professor Lewin goes thru this same derivation.

Unit 4 ==> Real Work Lifting Something

Don't we need to exert more force to lift a mass rather then just equal it's gravitational force?

More force but no more extra work

• In order to lift the bucket, we do in fact have to lift with a force greater than mg, otherwise there would be no net force and therefore no acceleration.
• Let's write this force as F=mg+K where K is the extra bit of force more than the weight force (can be any value greater than zero).
• Now we can imagine breaking down the work we do into two parts
• The work done by the part of our force that balances gravity (let's call it Wg) 
• The work done by the extra part of our force (let's call it WK).
• Considering only the work Wg, then this will naturally be just mg times the distance we lift.
• The extra force K does not go into fighting gravity, but instead into accelerating the bucket, which in turn increases its kinetic energy. So the work WK goes strictly into the bucket's kinetic energy.
• Now we assume that once we're done lifting the bucket, it's now at rest, and thus has no kinetic energy.
• But we know that the bucket should have kinetic energy equal to WK! 
• So this means that in order to bring the bucket to rest, we must have actually done some extra work to slow the bucket back down to rest. 
• And the only way to get rid of a kinetic energy of WK is to do work of −WK. 
• So these two works balance each other out
• And we get that the total work is just Wg. 
• So what this really means here is that the force must vary over the lifting. It has to start as greater than mg to get it moving, but it must end as less than mg to slow it down. This varying force means that the final work ends up being exactly Wg.  
• MIT Professor Lewin evaluates the net work done here at about 6 minutes into his lecture. He also does a demonstration of this activity.

Unit 4 - Problem 10 ==> Generalized Equations

Given:

• Acceleration of gravity  = 9.8 m/s/s
• Mass of the object = 1 kilogram
• Height over ground = 50 meters
• G = 6.674 x 10^-11
• Radius of the earth = 6400 kilometers = 6,400,000 meters
• Mass of the earth = 5.97 x 10^24 kilograms

Calculate: Gravitational potential energy using mgh

• $$ E_{potential} = m * g * h = 1 * 9.8 * 6400009 = 62720088.2  $$

Calculate: Potential energy using (G*M1*M2)/r

• $$ E_{potential} = G_{earth} * M_{object} * M_{earth} $$ 
• $$ E_{potential} = \left(\frac{6.674^{-11} * 1 * 5.97^{24}}{6400009}\right) $$
• $$ E_{potential} = 6.225581870275496 * 10^-6 * 10^{-11} * 10^{24} $$
• $$ E_{potential} = 6.225581870275496 * 10^7 $$
• $$ E_{potential} = 62255818.70275496 $$

Calculate: Difference & % difference in two calculations

• $$ E_{mgh} - E_{GM1M2/r} = difference$$
• $$ 62720088.2 - 62255818.70275496 = 464269.4972450435 $$
• $$ \left(\frac{difference}{E_{GM1M2/r}}\right) * 100 = percent_{difference}$$
• $$ \frac{464269.4972450435}{62255818.70275496} * 100 = 0.745744746305133$$

Unit 4 - Problem 8 ==> Challenge with a spring

Given:

• The mass of an object is compressed into the spring for a distance of delta x.
• mass = 500 grams =  0.5 kilograms
• delta x = 10 cm = 0.1 meters
• The spring  has a spring constant of 500 newtons/meter
• There is a resistance section:
• Whose length is 0.8 meters
• And which exerts a force of 1 newton
• The mass moves up a curving ramp that is 0.5 meters high
• Two questions
• Where does the mass eventually come to rest on  the resistance section as measured from the spring?
• What is the maximum height above the ramp achieved by the mass?

Calculate: Initial energy stored by the mass compressing the spring

• $$ E_{initial} = \frac{1}{2} k {\Delta{x}}^2 $$ 
• $$ E_{initial} = 0.5 * 500 * 0.1^2 = 2.5\;{joules} $$

Calculate: Energy dissipated traveling over the friction

• $$ E_{friction} = F_{friction} L_{friction} $$
• $$ E_{friction} = 1 * 0.8 = 0.8\;{joules} $$

Calculate: Cycles over the friction to dissipate the initial energy

• A full cycle entails two passes over the friction.
• $$ E_{friction-cycle} = 2 E_{friction} = 2 * 0.8 = 1.6\;{joules}$$
• $$ cycles = \frac{E_{initial}}{E_{friction-cycle}} = \frac{2.5}{1.6} =  1.5625 $$
• 1 full cycle  occurs leaving 2.5 - 1.6 = 0.9 joules of energy left
• Another right hand crossing occurs leaving 0.9 - 0.8 = 0.1 joules to come back and stop in the resistance area 
• $$ \frac{0.1\;{joules}}{0.8\;{friction-joules}} = \frac{x_{stop-from-right\;{m}}}{{0.8\;{m}}}$$
• There the mass stops 0.1 meters into the resistance from the right
• Or 0.7 meters from the spring

Calculate: Maximum energy at take-off of ramp

• Obviously, the maximum energy will occur with the minimum crossings of the friction and this occurs on the first cycle.
• $$ E_{take-off} = E_{initial} - E_{friction} - E_{up-ramp} $$
• $$ E_{take-off} = 2.5 - 0.8 - (m*g*h) $$
• $$ E_{take-off} = 2.5 - 0.8 - (.5*10*.5) $$ 
• $$ E_{take-off} = 2.5 - 0.8 - 2.5 = -0.8\;{joules}$$ 

Calculate: Maximum height above ramp

• The mass never takes off nor ever gets to the top of the ramp.

Unit 4 - Problem 7 ==> Basic Springs

Given:

• Mass of block = 5 kg
• Velocity of block = 2 m/s
,li>Spring constant = 1000

Calculate: Energy in moving block

• $$ E_{block} = \frac{1}{2} * m_{block} * V_{block}^2$$
• $$ E_{block} = 0.5 * 5 * 2^2 = 10_{joules}$$

Maximum deformation of spring results transfer of all energy to it.

• $$ E_{spring} = \frac{1}{2} * k * \Delta{x}^2 $$
• $$ E_{spring} = 10_{joules} = 0.5 * 1000 * \Delta{x}^2 $$
• $$ 10 = 500 * \Delta{x}^2 $$
• $$ \Delta{x}^2 = \frac{10}{500} = .02 $$
• $$ \Delta{x} = 0.1414213562373095 = 0.14 $$

Unit 4 - Problem 5 ==> Ski Jump

Calculate the energy of the jumper leaving the ramp

• $$ E_{potential} = m*g*h = m(10)(90) $$
• $$ E_{lost-on-jump} = m*g*h = m(10)(10) $$
• $$ E_{leaving- jump} = E_{potential} - E_{lost-on-jump} = 900m - 100m = 800m$$

Calculate the jump velocities both horizontal and vertical

• $$ E_{leaving-jump} = E_{kinetic-energy} = 0.5 *m*V_{0-jump}^2 $$
• $$ 800m = 0.5*m*V_{0-jump}^2 $$
• $$ V_{0-jump}^2 = 1600 $$
• $$ V_{0-jump} =  40.0$$
• $$ \frac{V_{0-y-jump}}{V_0} = \sin\alpha = 0.5$$
• $$ V_{0-y-jump} = V_0 * .5 $$ 
• $$ V_{0-y-jump} = 20.0$$
• $$ \frac{V_{0-x-jump}}{V_0} = \cos\alpha = 0.866$$
• $$ V_{0-x-jump} = V_0 * .866 $$
• $$ V_{0-x-jump} = 34.64$$

Calculate: The energy on landing and the landing velocities

• $$ E_{landing} = E_{kinetic} + E_{jump-height} $$
• $$ E_{landing} = 800m + m(10)(10) = 900m $$
• $$ 0.5*m*V_{landing}^2 = 900m $$
• $$ V_{landing}^2 = 1800 $$
• $$ V_{landing} = 42.42640687119285 $$
• The horizontal jump velocity is a constant.
• $$ V_{landing}^2 = V_{0-x-jump}^2 + V_{y-landing}^2$$
• $$ V_{landing}^2 - V_{0-x-jump}^2 = V_{y-landing}^2$$
• $$ V_{y-landing}^2 = 1800 - 1199.9296 = 1800 - 1200 = 600 $$
• $$ V_{y-landing}  = 24.49489742783178 $$

Calculate: Time to peak using vertical component of jump velocity

• $$ t_{jump-to-peak} = \frac{V_{peak} - V_{0-y-jump}}{a} $$
• $$ t_{jump-to-peak} = \frac{-20}{g} $$
• $$ t_{jump-to-peak} = 2.0$$

Calculate: Time from peak using the vertical component of the landing velocity

• $$ t_{peak-to-landing} = \frac{V_{peak} - V_{y-landing}}{a} $$
• $$ t_{peak-to-landing} = \frac{- V_{y-landing}}{a}$$
• $$ t_{peak-to-landing} = \frac{24.49489742783178}{g} = 2.449489742783178 $$

Calculate:Total time is sum of time up to peak plus time down to ground

• $$ t_{total} = t_{jump-to-peak} + t_{peak-to-landing}$$
• $$ t_{total} = 2.0 + 2.449489742783178 = 4.449489742783178 $$

Calculate d based on total time and horizontal velocity at jump

• $$ d_{total} = V_{0-x-jump} * t_{total} $$
• $$ d_{total} = 34.64 * 4.449489742783178 $$
• $$ d_{total} = 154.13032469000927 = 154 $$

Unit 4 - Problem 4 ==>Maximum car speed

How fast given horsepower and air resistance?

• Resistance in joules is: $$R_{air} = b*v^2 = .1{V^2}$$
• Power consumed is: $$ P_{resistance} = R * V $$
• Because:  $$F_{resistance} \frac{D}{t} = \frac{(F_{resistance}*D)}{t} = \frac{W}{t} = P$$
• Car has 150 hp where 1 hp = 746 watts
• Power of the car in watts:$$ W_{car} = 150 * 746 = 111,900 $$
• Top speed is when power of resistance equals the power of the car: $$.1{V^3} = 150 * 746  $$
• $$ V^3 = 1,119,000 $$
• In meters/second: $$ V_{max} = 103.8 $$

Unit 4 - Problem 3 ==> Weight lifting workout

W = m*g*h = work done in joules

• m = 20, g = 10 m/s/s, h = 1 m ==> for one repetition
• $$W = 20*10*1 = 200$$
• Workout is for 4 minutes which equals 240 seconds
• Every second we do two-thirds of a repetition
• $$Repetitions = (\frac{2}{3}) 240 = 160 $$
• $$joules = 200 * 160 = 32000 $$
• There are 4200 joules in a calorie.
• $$calories = \frac{32000.0}{4200.0} = 7.6 $$

Unit 4 - Problem 2 ==> Rollercoaster with friction

Given

• Similar to Unit 4 - Problem 2
• Initial height = 45 meters
• There is a 20 meter section before the loop that creates a friction force.
• What is maximum value of friction force?

Rationale

• $$E_{potential} = E_{top-of-loop} + E_{kinetic-top-of-loop}$$
• $$m_{car} a_{gravity} h = (m_{car} a_{gravity} h_{top-of-loop}) + (\frac{1}{2} m_{car} V^2_{top-of-loop}) - (F_{x-friction-max}  d_{bad-track}) $$

Calculations

• $$(200*45*10) = (200*20*10) + (\frac{1}{2}*200*10^2) - (F_{x-friction-max}*20)$$
• $$90000 = 40000 + 10000 - (F_{x-friction-max}*20) $$
• $$90000 - 50000 = -20F_{x-friction-max}$$
• $$40000 =  -20F_{x-friction-max}$$
• $$F_{x-friction} = -2000 $$

Unit 4 - Work, Energy, and Power

Definition of Work

• A scalar quantity that can be described as the product of a force and the distance through which it acts in the direction of the force.
• The SI unit of work is the joule (J)
• $$W_x = F_x * d_x$$
• From Wikipedia

Definition of Energy

• The kinetic energy of an object is the energy which it possesses due to its motion
• The SI unit of work is the joule (J)
• $$E_{kinetic} = \frac{1}{2}mv^2$$
• From Wikipedia

Definition of Potential Energy

• The potential energy is the energy of a body or a system due to the position of the body or the arrangement of the particles of the system
• The SI unit of work is the joule (J)
• $$U_{gravity} = mgh$$

Conservation of Energy

• $$E_{final} = E_{initial} - E_{lost}$$
• Losses can be due to friction and the heat generated.

Definition of Power

• Power is the rate at which energy is transferred, used, or transformed
• The SI unit of power is the watt (W), which is equal to one joule per second
• $$P = \frac{W}{\Delta{t}}$$
• From Wikipedia

Friction and Energy Dissipated

• Friction is the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other
• From Wikipedia

General Equations of Force & Potential Energy

Dimension	Small Distances	Large Distances
Force	mg	$(G*M1*M2)/r^2$
Potential Energy	mgh	$(G*M1*M2)/r$

Conservative Force vs Force

• If we have a conservative force, then (PE-a + KE-a) = (PE-b + KE-b)
• PE + KE = Mechanical Energy
• Gravity is a conservative force as is a spring force.
• Friction is not a conservative force because it is dependent on the route taken.
• One route could be longer than another resulting in more frictional work.
• From MIT Professor Lewin
• Useful constants with General Gravity equation
• G constant = 6.674 * 10^-11
• Radius  of the earth = 6.4 * 10^6 meters
• Mass of the earth = 5.97 * 10^24 kilograms
• Also in another lecture by MIT Professor Lewin, he takes up the subject of Power and our ongoing "Energy Crisis".

Unit 4 - Problem 1 ==> Rollercoaster Design

Given

• A rollercoaster car with mass m at height h goes into a loop.
• At the top of the loop of radius r the car must go a velocity of V to be safe
• What is the minimum height the car must start at?

Rationale

• $$E_{potential} = E_{top-of-loop} + E_{kinetic-top-of-loop}$$
• $$m_{car} a_{gravity} h_x = (m_{car} a_{gravity} h_{top-of-loop}) + (\frac{1}{2} m_{car} V^2_{top-of-loop}) $$
• $$a_{gravity} h_x = (a_{gravity} h_{top-of-loop}) + (\frac{1}{2} V^2_{top-of-loop}) $$

Calculations

• $$10h_x = (10*20) + (\frac{1}{2}*10^2)$$
• $$10h_x = 200 + 50 = 250$$
• $$h_x = 25$$

Height vs Loop Radius

• Derivation from MIT Professor Lewin at about 28:40 on the video.
• At any point along the track, $$ mgh = mgy + \frac{1}{2}m v^2 $$
• Which simplifies to, $$ 2g (h-y) = V^2$$
• At the lop of the loop, $$ y = 2r $$
• From previous work, at the apex of the loop,  $$ a_{centripetal} = g = \frac{V^2}{R}$$
• $$ V^2 = g * R $$
• Therefore, $$ 2g*(h-2r) \ge g*r $$
• $$ 2h - 4r \ge r $$
• $$ h  \ge \frac{5}{2} r $$
• Regardless of mass of object, for it to stay on the track, it must originate from a height of at least 2.5 times the radius of the loop.

anas

Here is what you posted:" Hello everybody. While I was trying to solve this problem,I found a contradiction,and that confuses me because all the equations that I used are right.well,let me clarify: First of all,I calculated the Force of the air = bV^2=410^-5 Then I calculated the weight of the feather,AKA Fmg=mg=0,01 After that,until I know the final force applied on it,I calculated the difference: P-F air=9,9610^-3 Okay,then,until I get the acceleration value,I used the Newton equation : F=ma to figure it out,I of course used the final force (9,9610^-3),but and here's the surprise,I find the value of acceleration :9,96 m/s^2,and that just blows my mind up ! It's a too high value and,as described earlier in the course,the feather fall very slowly (almost at a constant speed). Could you please clarify this for me,show me if there's a mistake in my measurement. P.S:I couldn't find the speed using only the acceleration that I've calculated,I tried to use the system that was given as a solution in the unit,but in vain,there's a lot of unknown variables. Thank you."
Now I am trying to figure out what you did and I have some questions:

What you wrote	A question	Comments
Force of the air = bV^2=410^-5	We don't know V or V^2. I don't understand where you got 410^-5.	We know only: Weight of feather = .001 kg Feather resistance equation of bV^2 b = .04
Then I calculated the weight of the feather,AKA Fmg=mg=0,01	The weight of the feather is a given. It is .001 kg	No calculation is necessary
P-F air=9,9610^-3	I don't know what this is a calculation of or how you did the calculation
I find the value of acceleration :9,96 m/s^2	The acceleration down is gravity at 10 m/s/s. What was it that surprised you?
Could you please clarify this for me,show me if there's a mistake in my measurement.	For me to do this, I need to see your calculations.	I need more data from you.
I couldn't find the speed using only the acceleration that I've calculated	The problem is to express the forces up and down; set them equal; and solve for V.	This is what I did in my posted answer.
There's a lot of unknown variables		You have over achieved and been stumped. Your approach did not work out for you. Time to try another.

Unit 3 - Problem 8 => Sky diving & the earth

Given:

• 10 minute sky dive
• Mass of sky diver is 65 kg
• Mass of the earth $6\times10^{24}$

Reasoning:

• Since opposing forces must always be equal (for bodies at rest or constant velocity), then the force on the sky diver must also be acting on the earth.
• Note the time is 10 minutes, there for or calculations must use $(10\;minutes)\times60\;seconds = 600 seconds$.

Calculations:

$$F_{skydiver}=F_{earth}$$
$$F_{skydiver}=m_{skydiver}*a_{skydiver}=65\;{kg}*10\;{\frac{m}{s^2}}=650\;N$$
$$F_{earth}={6*10^{24}}*a_{earth}=650\;N$$
$$a_{earth}=\frac{650}{6*10^{24}}$$
$$x_{earth}=\frac{1}{2}*a_{earth}*t^2$$
$$x_{earth}=(.5*\left(\frac{650}{6*10^{24}}\right)*{600}^2$$
$$x_{earth}=195*10^{-19}=1.95*10^{-17}$$

Unit 3 - Problem 6 - More Inclined Plane

Given:

• When $mass_1$ is moving up, $a$ is positive
• Observation: This is an Atwood Machine variation using an inclined plane on one side
• Don't solve the details.
• Use your intuition.

Reasoning (aka Intuition)

• When $mass_2$ is moving down, $g$ is negative since up is positive and therefore down is negative.
• $mass_2\times\sin\alpha = mass_{2-effective}$ due to previous problem where we calculated the inclined slope effective force.
• Consider the case where the numerator is $(M_2\times\sin{\alpha}-M_1)$ and the denominator is $M_1+M_2$
  • Suppose $M_2 = 0$ then $a=g\left(\frac{-M_1}{M_1}]\right)=-g$ which we know is TRUE.
  • Suppose $M_1 = 0$ then $a=g\left(\frac{M_2\sin(\alpha)}{M_2}\right)=g\times\sin(\alpha)$ which we know is TRUE.
Clearly, if the subtraction and addition were reversed, the results would be FALSE since the other way is TRUE.
• Also, no other arithmatic operation would produce these simple results and therefore must be FALSE.

Unit 3 - Problem 5 => San Francisco Parking

Given:

• Friction between tires and road prevents car from slipping.
• $F_f$ has a max value of $80\%$ of car weight.
• The maximum angle is $\alpha$.

Reasoning

• $F_{max} = 0.80 * W_{car}$. Beyond $F_{max}$ the car slides down the hill.
• $F_{parallel} = W_{car} * \sin\alpha$ (obtained from previous problem)is the actual force on the roadway.
• The maximum angle occurs when the $W_{car} * \sin\alpha = 0.80 * W_{car}$

Calculations:

$$\sin\alpha = \left(\frac{.8W_{car}}{W_{car}}\right)$$
$$\sin\alpha = .8 $$
$$\alpha = \arcsin{.8} = 53.13^{\circ}$$

 

Some observaions

 

I do not ever want to be on a grade of 53%. I could not stand up on it. Consider the following:

• The famous Lombard Street in San Francisco, the curviest street in the world only has a grade of 14.3 %.
• The steepest street in San Francisco is Filbert Street between Hyde and Leavenworth at a grade of 31.5 %.
• An interstate is out of standard if it has a grade > 7%.
• Local roads are much higher. 12% or 15% are sometimes allowed.
• Driveways can be as much as 30% for a short distance
• Railroads generally avoid grades over 2%.
• Even commercial airplanes don't have numbers as high as 53%,
• A330, out of maintenance - empty, engineers asked for a full power takeoff. Full back stick after normal rotation to stop speed going silly, settled at about 28 degrees, 8,500 f.p.m climb
• 747 stick shaker starts at 17-17.5 deg, so we never intentionally go above that.

The concept of slope (slope is normally described by the ratio of the "rise" divided by the "run" between two points on a line. ) applies directly to grades or gradients in geography and civil engineering. Through trigonometry, the grade m of a road is related to its angle of incline theta.

$${incline} = \tan\theta$$ $$\theta = \arctan{(incline)}$$

Unit 3 - Problem 4 =>Inclined Planes Reducing Gravity

Rational:

• The normal force resulting from mass times gravity is broken up into a parallel force and a perpendicular force.
• From geometry, we can see that the inclined plane angle alpha is also the angle closest to the mass.
• Further the parallel force is the equivalent of gravity since it is the force acting to pull the mass down the inclined plane.
• The effective acceleration down the inclined plane is a function of the parallel force.
• Thus it is shown that the effective acceleration is always less than or equal to normal gravity depending upon the angle alpha.
• $\sin{\alpha}=\frac{opposite}{hypotenuse}=\frac{F_{parallel}}{m*g}$
• $F_{parallel}=m * g * \sin\alpha=F_{normal}$
• $\cos{\alpha}=\frac{adjacent}{hypotenuse}=\frac{F_{perpendicular}}{m*g}$
• $F_{perpendicular}=m * g * \cos\alpha$
• $m*a_{effective}=F_{parallel}=m * g * \sin\alpha$
• $a_{effective}=g*\sin\alpha$$

Unit 3 - Problem 3 => Gravity on Mt. Everest

Given:

• $g_{sea-level} = \frac{10\;meters}{seconds^2}$
• Gravity is inversely proportional to the square of the distance from the center of the earth.
• The earth's radius is aproximately 6,400 km.
• Mt. Everest height is about 9 km.
• Calculate constant of proportionality k.
• Calculate by what percentage does the acceleration due to gravity change when we stand on Mt. Everest.

Rational for constant of proportionality:

• $g = \frac{k}{r^2}$

Calculating the constant of proportionality:

• $10 = \frac{k}{6400^2}$
• $k = 10*6400^2$

Rational for finding the acceleration due to gravity when we stand on Mt. Everest.

• Mt. Everest is about 9 km high so its distance from the center of the earth is about 6409 km.
• $g_{Mt.Everest} = \frac{k}{r^2} = \frac{10*6400^2}{r^2}$

Calculating Mt.Everest force of gravity

• $g_{Mt.Everest} = \frac{10*6400^2}{r^2}$
• $g_{Mt.Everest} = \frac{10*6400^2}{6409^2} = 9.971934215130506 = 9.97$

Rational for percent difference:

• ${percent-difference} = 100 \times \left(\frac{Difference}{{Comparison Value}}\right)$

Calculating the % difference in force of gravity at sea level vs Mt.Everest

• ${percent-difference} = 100 \times \left(\frac{g_{sea-level}-g_{Mr.Everest}}{g_{sea-level}}\right)$
• ${percent-difference} = 100*\left(\frac{(10-9.97)}{10}\right)= 0.2999999999999936 = 0.3\;\%$

Unit 2 - Problem 2 => Man in elevator

The man weighs 800 newtons and wants to weigh 600 newtons. Find the man's mass first where gravity is 10 m/s/s. $$F = m*a$$ $$m = \frac{F}{a} = \frac{800}{10} = 80\;{kilograms} $$ Now find the acceleration that would make this mass weigh 600 newtons. $$a = \frac{F}{m} = \frac{600}{80} = 7.5\frac{m}{s^2}$$ The downward acceleration to reduce the overall acceleration is: $$ 7.5 = g - a_{down}$$ $$a_{down} = g - 7.5 = 10 - 7.5 = 2.5 \frac{m}{s^2}$$ The distance down traveled in 5 seconds, assuming starting from a dead stop, is given by: $$y_{down} = \frac{1}{2} a t^2 = .5 * 2.5 * (5*5) = 31.25 m$$

Unit 3 - Problem 1 => Feather Terminal Velocity

Given:

$$weight_of_feather = .001kg$$ $$feather_air_resistance = b * V^2$$ $$b = .04$$

Reasoning:

$$F_{down} = m * a = .001kg * 10\frac{m}{s^2} = .01{Newtons}$$ $$F_{up} = b * V^2 = .04V^2$$ When these two force are equal, there is no further force and the V at that point is the terminal velocity. $${(F_{up} = b * V^2 = .04V^2)} = {(F_{down} = m * a = .001kg * 10\frac{m}{s^2} = .01{Newtons})}$$

Solving:

$$.04V^2 = .01{Newtons}$$ $$V^2 = \frac{.01}{.04} = \frac{1}{4}$$ $$V = \frac{1}{2} = 0.5\frac{m}{s^2}$$

Unit 3 - Atwood Machine

A pulley with two weights(masses): m1 and m2

The weights are connected so their accelerations are identical and given by this equation:

$a = g*\left(\frac{m2 - m1}{m2+m1}\right)$

Unit 2 - Challenge Question - Math Symbolics

Given:

• Use acceleration of gravity up is negative the acceleration of gravity down which is equal to 10 meters/sec/sec
• The horizontal distance from the kicking point at the top of a hill to the spot where the ball finally impacts further down the hill is $$ \Delta{x} $$
• Drawing a horizontal line at the point of kicking:
• The angle of the kick up is $$ \alpha_{up} $$
• The hill down is a straight inclined plane
• The angle of the hill down from the horizontal is $$ \beta $$
• The initial velocity of the kicked ball is $$ V_0 $$

Rationale:

• The two given angles are used differently:
• Angle alpha is used to determine the x and y initial velocities
• Transversing angle beta between two parallel lines allows angle beta to be the right most angle of a right triangle. This triangle has a y-axis equal to the hill height andd an x-axis equal to Delta-x.
• Two simultaneous equations of the height of the hill allow handling two unknowns in each of these equations.

Solve: Initial velocities of the kicked ball

• $$ \frac{V_{0-y}}{V_0} = \sin{\alpha} $$
• $$ V_{0-y} = \sin{\alpha} * V_0$$
• $$ \frac{V_{0-x}}{V_0} = \cos{\alpha} $$
• $$V_{0-x} = \cos{\alpha} * V_0 $$

Solve: Time for kicked ball to go Delta x distance

• $$ \Delta{x} = (V_{0-x})t $$
• $$ \Delta{x} = (\cos{\alpha} * V_0)t $$
• $$ t = \frac{\Delta{x}}{(\cos{\alpha}V_0)} $$

Solve: Height of hill - 1st way

• $$ \frac{-y_{hill}}{\Delta{x}} = \tan\beta $$
• $$ y_{hill} = -(\Delta{x})\tan\beta $$

Solve: Height of hill - 2nd way

• $$ y_{hill} = (V_{0-y})t + \frac{1}{2}{g}{t^2} $$
• $$ y_{hill} = (\sin{\alpha} * V_0)t + \frac{1}{2}{g}{t^2} $$
• $$ y_{hill} = (\sin\alpha * V_0)\left(\frac{\Delta{x}}{(\cos{\alpha}V_0)}\right) + \frac{1}{2}{g}\left(\frac{\Delta{x}}{(\cos\alpha V_0)^2}\right) $$
• $$ y_{hill} = \left(\frac{\sin\alpha}{\cos\alpha}\right)\Delta{x} - \left(\frac{g}{2}\right)\left(\frac{\Delta{x}^2}{(V_0\cos\alpha)^2}\right) $$
• $$ \left(\frac{\sin\alpha}{\cos\alpha}\right) = \left(\frac{opposite}{hypotenuse}\right)\left(\frac{hypotenuse}{adjacent}\right) = \tan\alpha $$
• $$ y_{hill} = \tan\alpha\Delta{x} - \left(\frac{g}{2}\right)\left(\frac{\Delta{x}^2}{(V_0\cos\alpha)^2}\right) $$

Solve the two height of hill equations together

• $$ y_{hill} = y_{hill} $$
• $$ -(\Delta{x})\tan\beta = \tan\alpha\Delta{x} - \left(\frac{g}{2}\right)\left(\frac{\Delta{x}^2}{(V_0\cos\alpha)^2}\right) $$
• $$ -(\Delta{x})\tan\beta = \tan\alpha\Delta{x} - \left(\frac{g}{2}\right)\left(\frac{\Delta{x}^2}{(V_0\cos\alpha)^2}\right) $$
• $$\tan\alpha\Delta{x} + (\Delta{x})\tan\beta = \left(\frac{g}{2}\right)\left(\frac{\Delta{x}^2}{(V_0\cos\alpha)^2}\right)$$
• $$\Delta{x}(\tan\alpha + \tan\beta) = \left(\frac{g}{2}\right)\left(\frac{\Delta{x}^2}{(V_0\cos\alpha)^2}\right)$$
• $$(\tan\alpha + \tan\beta) = \left(\frac{g}{2}\right)\left(\frac{\Delta{x}}{(V_0\cos\alpha)^2}\right)$$
• $$\Delta{x} = \left(\frac{2(\tan\alpha + \tan\beta)(V_0\cos\alpha)^2}{g}\right) $$

Adding LaTeX to My Blog

Using LaTeX in Blogger and rendering it with Mozilla Firefox requires insertion of the following HTML after the header (<head>) in the Blogger template (Design→Edit HTML→Edit Template).

From the TEX exchange, to enable MathJax (aka LaTeX), just drop in:

<script type="text/javascript" 
src="http://cdn.mathjax.org/mathjax/latest/MathJax.js">
MathJax.Hub.Config({
 extensions: ["tex2jax.js","TeX/AMSmath.js","TeX/AMSsymbols.js"],
 jax: ["input/TeX", "output/HTML-CSS"],
 tex2jax: {
     inlineMath: [ ['$','$'], ["\\(","\\)"] ],
     displayMath: [ ['$$','$$'], ["\\[","\\]"] ],
 },
 "HTML-CSS": { availableFonts: ["TeX"] }
});
</script>

To prove that this works, consider an ordinary presentation of Einstein's famous equation:



E = m * c^2 
Now consider an LaTeX presentation: $$E = m * c^2$$

Unit 2 - Problem 8 Dimensional Analysis

 

Definition

• Dimensional Analysis is the process of determining the M (mass), L (length), and T (time) exponents for a given variable.
• For example, velocity is equal to length raised to the power of one divided by time raised to the power of one. Mass is not involved so it's exponent is zero making it's value 1 and thus not reported. Consequently, we have: $$V = \frac{length}{time} = \frac{L^1}{T^1} = L^1 * T^-1$$
• For an MLT value = 1,-1

Unit 2 - Problem 8 asks:

• What is the MLT for:$$\frac{h^2}{ m x^2} = E  = m c^2$$
• So solving for h:
• $$h^2 = (m * x^2) * m * c^2$$
• $$h^2 = m^2 * x^2 * c^2$$
• $$h = m * x * c$$
• Note that: x = L(ength); and c(velocity, i.e. speed of light) = L/T
• Which in MLT I think means:
• $$MLT = M^1;L^1;(\frac{L}{T})^1 = M^1;L^1;(L^1,T^−1)$$
• $$MLT = M^1;L^2:T^−1$$
• Making MLT parameters = 1,2,-1

Another use for MLT

• Is shown by MIT Professor Lewin in this video around 22 minutes
• Units, Dimensions, and Scaling Arguments

Unit 2 - Problem 12 - Minimum Angle to Hit Ship

We know the shape of the flight of a cannon ball is a parabola and the vertex form of parabola equation is ( Source = Math Warehouse ):

• $$ y = -a(x-h)^2 + k$$

We know three cases so we can  solve for the three unknown variables a, h, and k.:

• $$x= 0 , y=0$$
• $$x=100, y=40$$
• $$x=200, y=0$$

We can substitute our knowns:

• $$0 = -a*h^2 + k$$
• $$40 = -a*(100-h)^2 + k$$
• $$0 = -a*(200-h)^2 + k$$

Solve for definition of k:

• $$k = a*h^2$$
• $$k = -.004 * 100^2$$
• $$k = -.004 * 10000$$
• $$k = 40.0$$

Substitute our knowns:

• $$40 = -a*(110-h)^2 + a*h^2$$
• $$0 = -a*(200-h)^2 + a*h^2$$

Solve for h:

• $$0 = -a((200-h)^2 - h^2)$$
  • $$0 = (200-h)^2 – h^2$$
  • $$h^2 = (200-h)^2$$
  • $$h = 200-h$$
  • $$2h = 200$$
  • $$h = 100$$

Solve for a:

• $$40 = -a((100-h)^2 – h^2)$$
  • $$40/((100-h)^2 – h^2) = -a$$
  • $$a = -40/((100-h)^2 – h^2)$$
  • $$a = -40/((100-100)^2 - h^2)$$
  • $$a = -40/(-10000)$$
  • $$a = -40/(-10000) = 4/10000$$
  • $$a = .004$$

My parabola's equation is:

• $$ y = -.004(x-100)^2+40$$

To get initial velocity we need an initial angle.So let's calculate where y is when x = 1 meter.

$$y = -.004(1-100)^2 +40$$

$$y = -.004(9801) + 40$$

$$y = -39.204 + 40 = 0.796$$

$$opp/adj = 0.796/1 = \tan{angle}$$

$$arctan (0.796) = 38.5197895 degrees$$

We know: V² = V0² + 2aΔx

• V = 0 at peak of parabola
• a = 10
• x = 100
• 0 = V0² + 2*10*100
• VO^2 = 2000
• VO = 44.72

Unfortunately the iceberg is between  110 m (as given by the problem) and 115 m (per previous calculations) from iceberg and not at center of parabola (100m). Somehow a higher initial velocity is required to move the parabolic center to the 40 m level of the iceberg.

I don't know how to calculate this problem from here and will do it by trial and error observation.

My best answer is 38.91553 degrees at 45 meters/sec. You can see this is slightly higher values for both initial angle and velocity.

With a real cannon, the initial velocity would be determined by the gunpowder load that would only come in fixed increments. But that would be a different problem.