# Mechanics Homework, Set 1

1. A race car makes two and a half laps around a circular track in 3.7 MIN. What is the car’s average angular speed? Express your answer using two significant figures.

2. A wheel of radius 1.2 M rotates at a uniform speed. If a point on the rim of the wheel has a centripetal acceleration of 2.8 m/s s, what is the point’s tangential speed? Express your answer using two significant figures.

3. A jet pilot puts an aircraft with a constant speed into a vertical circular loop. If the speed of the aircraft is 680 km/h and the radius of the circle is 2.0 km, calculate the normal force exerted on the seat by the pilot at the bottom of the loop. Express your answer in terms of the pilot’s weight W. B) Calculate the normal force exerted on the seat by the pilot at the top of the loop. Express your answer in terms of the pilot’s weight.

4. A pendulum swinging in a circular arc under the influence of gravity, as shown in the figure , has both centripetal and tangential components of acceleration. If the pendulum bob has a speed of 2.6 m/s when the cord makes an angle = 16 degrees with the vertical, what are the magnitudes of the components at this time? Express your answers using two significant figures separated by a comma. B)Where is the centripetal acceleration a maximum? C) What is the value of the tangential acceleration at that location?

5. During the Apollo lunar explorations of the late 1960s and early 1970s, the main section of the spaceship remained in orbit about the Moon with one astronaut in it while the other two astronauts descended to the surface in the landing module. Assume the mass of this section is 5000 kg. If the main section orbited about 46 mi above the lunar surface, determine the gravitational potential energy. Express your answer using two significant figures.B) Determine the total energy. Express your answer using two significant figures.

Express your answer using two significant figures.

Answers

1. $(angular-speed) = (total-angle)/time =2*\pi*2.5/(3.7*60) =0.07 (rad/sec)$

2. let $\omega$ be the angular speed. Then the acceleration is

$a =\omega^2*R$

$\omega = \sqrt{a/R}$

speed $v = \omega*R =R*\sqrt{a/R} =1.83 m/s$

3. The centrifugal force is $F_{cf}= m*a =m*\omega^2*R$ where $\omega$ is the angular speed

a)

At the bottom of the loop the reaction of the seat is $R = m*g +m*\omega^2*R$

$R/G = 1 +\omega^2*(R/g) = 1+ (v/R)^2*(R/g)$

$680 km/h = 680000/3600 =188.89 m/s$

$R/G = 1+188.89^2/2000/9.81 =2.82$

b)

At the top of the loop the reaction of the seat is $R = -m*g +m*\omega^2*R$

$R/G = -1 +\omega^2*R/g = -1+ (v/R)^2*R/g$

$R/G = -1+188.89^2/2000/9.81 =0.82$

4.The speed is always tangential to trajectory. Therefore the centrifugal acceleration is

$a_cf =\omega^2*R + g*cos(\theta)$

where $\omega = v/R$

$a_{cf} = v^2/R +g*cos(\theta) = 2.6*2.6/0.75 +9.81*cos(16) =18.44 m/s/s$

the tangential acceleration is simply

$a_t = g*sin(\theta) = 9.81*sin(16) =2.70 m/s/s$

B)

the centripetal acceleration is maximum where the tangential speed is maximum and this is happening when the angle $\theta =0$ (the pendulum is vertical).

C)

At the point where the pendulum is vertical the tangential acceleration is null. $a_t = 0$

5. One needs to know the value of g (gravitational acceleration) at the surface of the moon.

$g_{moon} = k*M_{moon}/R_{moon}^2 = 6.67*10^-11 *7.35*10^22/1737100^2 =1.625 m/s^2$

$46 miles = 46*1609 = 74104 m$

A)

The potential energy at 46 miles is

$E_p =m*g*h =5000*1.625*74104 =601363750 J =0.60 GJ$

B)

the speed necessary to orbit the moon comes from equaling the centrifugal force with the weight

$m*v^2/R =m*g_{moon}$

$v^2 =g_{moon}*R = 1.625*(R_{moon}+R) =1.625*(1737100 +74100) =2943200 m^2/s^2$

the kinetic energy is $E_c = m*V^2/2 =5000*2943200/2 =7.358*10^9 J = 7.358 GJ$

The total energy

$E = E_p+E_c = (0.6+7.358)*10^9 =7.958*10^9 J = 7.958 GJ$

C)

the energy needed to escape the moon is equal to the kinetic energy of the main section when having the escape velocity. The escape velocity is the velocity when the gravitational force is equal to the centrifugal force at the surface of the moon.

$m*v^2/R =m*g_{moon}$

$v^2 =g_{moon}*R = 1.625*R_{moon} =1.625*1737100 = 2.82*10^6m^2/s^2$

The kinetic energy is $E_c = m*V^2/2 =5000*2.82*10^6/2 =7.06*10^9 J = 7.06 GJ$