PHYSICS 116A Prof. Ilias Perakis

Test # 3, Chapters 10-12 from Serway

$${\bf v} = {\bf v_{0}} + {\bf a} t \;\;,\;\; {\bf r} = {\bf... ...\ , \ \rm{\upsilon^{2} = \upsilon_{0}^{2} + 2 a (x - x_{0}) }$$

$$\rm{a_{r} = \frac{\upsilon^{2}}{r} \;\;,\;\; a_{t} = \frac{... ... \;\;,\;\; T=\frac{2 \pi r}{\upsilon}} \;\;,\;\; g=9.8m/s^{2}$$

$$\rm{a x^{2} + bx +c =0 \rightarrow x=\frac{-b \pm \sqrt{b^{2} - 4 a c}}{2a} } \ , \ G=6.67 \times 10^{-11} N m^{2}/kg^{2}$$

$$\rm{\underline{{\bf Friction}:} } \ \ f_{s} \le \mu_{s} N \ ... ...u_{k} N \ ; \ \rm{\underline{{\bf Spring}:}} \ \ F_{s} = -k x$$

$$K_{f} + \sum U_{f} = K_{i} + \sum U_{i} \ + \ W^{\rm{friction}}_{i \rightarrow f} \ + \ W^{\rm{ext}}_{i \rightarrow f}$$

$$K = \frac{1}{2} m \upsilon^{2} \ , \ U_{g} = m g y \ \rm{or}... ...= - \frac{G m_{1} m_{2}}{r} \ , \ U_{s} = \frac{1}{2} k x^{2}$$

$$W= {\bf F} \cdot {\bf d} = F d \cos \theta = \int_{x_{i}}^{x... ...x) dx \ , \ P(t) = \frac{d W(t)}{dt} = {\bf F} \cdot {\bf v}$$

$${\bf p} = m {\bf v} \ , \ {\bf I} = \int_{t_{i}}^{t_{f}} {\bf F} dt = {\bf \bar{F}} ( t_{f} - t_{i}) =\Delta {\bf p}$$

$$v_{1f} = \frac{m_{1} - m_{2}}{m_{1} + m_{2}} v_{1i} + \frac... ...+ m_{2}} v_{1i} + \frac{m_{2} - m_{1}}{m_{1} + m_{2}} v_{2i}$$

$${\bf r_{c}} = \frac{ \sum_{i} m_{i} {\bf r_{i}}}{\sum_{i} m_... ... {\bf F}_{\rm{ext}} = M \bf{a_{c}} =\frac{d {\bf p_{c}}}{dt}$$

$$\Delta s = r \ \Delta \theta \ , \ v = \omega \ r \ , \ I = \sum_{i} m_{i} r_{i}^{2} \ , \ K = \frac{1}{2} I \omega^{2}$$

$$\omega = \omega_{0} + \alpha t \ , \ \theta = \theta_{0} + ... ... \omega^{2} = \omega_{0}^{2} + 2 \alpha (\theta - \theta_{0})$$

$${\bf C} = {\bf A} \times {\bf B} \ , \ C = A B \sin \theta$$

$${\bf A} \times {\bf B} = - {\bf B} \times {\bf A} \ , \ {\bf A} \times {\bf A} =0$$

$${\bf i} \times {\bf j} = {\bf k} \ , \ {\bf j} \times {\bf k} = {\bf i} \ , \ {\bf k} \times {\bf i} = {\bf j}$$

$${\bf \tau} = {\bf r} \times {\bf F} \ , \ \tau_{\rm{net}} = I \alpha$$

$${\bf L} = {\bf r} \times {\bf p} \ , \ L = I \omega \ , \ \tau_{\rm{ext}} = \frac{ d {\bf L}}{dt}$$

$$F=G \frac{m_{1} m_{2}}{r^{2}} \ , \ G=6.67 \times 10^{-11} N m^{2}/kg^{2}$$

$$\rm{Earth \ Radius}= 6.37 \times 10^{6} \ m \ , \ \rm{Earth \ Mass}=5.98 \times 10^{24} \ kg$$

$$x(t) = A \ \cos( \omega t + \phi)$$

$$f = 1/T \ , \ \omega = \sqrt{\frac{k}{m}} \ , \ \omega = \s... ...c{g}{L}} \ , \ \omega = 2 \pi f \ , \ E = \frac{1}{2} k A^{2}$$

Multiple-Choice Problems

1.

A gentleman bug sits at the outer edge of a merry-go-round (distance R from the rotation axis) when he notices a lady bug enjoying the sun at a distance R/3 from the rotation axis. What is the lady bug's angular acceleration?
(A) three times the gentleman bug's (B) one third the gentleman bug's (C)$\star$ the same as the gentleman bug's (D) Impossible to know (E) Twice the gentleman bug's

cm Since the merry-go-round rotates as a rigid body, each of its points (as well as the bug sitting at any point) has the same angular displacement, angular velocity and angular acceleration.

2.

In the above question, if the gentleman bug's speed is 3 m/s, what is the speed of the lady bug?
(A) three times the gentleman bug's (B)$\star$ one third the gentleman bug's (C) the same as the gentleman bug's (D) Impossible to know (E) Twice the gentleman bug's

The linear speed of the gentleman bug can be expressed in terms of the angular velocity and its distance from the rotation axis: $v=\omega R$. Thus the angular speed of the merry-go-round is $\omega = v/R$. Since the lady bug has the same angular speed but sits at a distance R/3 from the rotation axis, its linear speed is:

$$v' = \omega \frac{R}{3}=\frac{v}{R} \frac{R}{3} = \frac{v}{3}=1m/s$$

3.

Stars originate as large bodies of slowly rotating gas, which slowly decrease in size due to gravity. As the star shrinks, its angular velocity
(A)$\star$ Increases (B) Decreases (C) Remains the same

(D) Impossible to tell (E) Oscillates

As the star's size decreases its moment of inertia also decreases according to the expression $I=\Sigma m_i r_i^2$. Since other celestial objects are very far away, practically, there is no external force (torque) acting on the star. Therefore, its angular momentum is conserved:

$$\tau=\frac{dL}{dt}=0 \quad \rightarrow \quad L=I \omega=constant.$$

That means the angular velocity must increase while its moment of inertia decreases to maintain the same angular momentum.

4.

You have two wheels with fixed hubs which weigh the same. The radius of the second wheel is three times that of the first. The mass of the hubs and spokes is small, so the total mass can be assumed to be uniformly distributed at a distance equal to the radius of the wheel. You apply the same force to both wheels, as shown in the figure, and start them rotating from rest. The angular acceleration of the second wheel is then
(A) three times that of the first (B)$\star$ one third that of the first (C) same as that of the first (D) nine times that of the first (E) one ninth that of the first

The moment of inertia of the second wheel is nine times as large as that of the first wheel, because I=MR² for wheels. The torque exerted on the second wheel is three times larger than the torque acting on the first one, because $\tau=Fd$, where F is the force and d is the moment arm which is equal to the radius of the wheels in our case (the force acts along the circumference). Then the angular acceleration of the second wheel is one third that of the first:

$$\alpha_1=\frac{\tau_1}{I_1}$$

and

$$\alpha_2=\frac{\tau_2}{I_2}=\frac{3\tau_1}{9I_1}=\frac{1}{3}\frac{\tau_1}{I_1}= \frac{1}{3}\alpha_1.$$

5.

Planet X has the same mass and one half the radius of the Earth. If the Earth's acceleration of gravity is g, what is it on planet X's surface?
(A) g/4 (B) g/2

(C) g (D) 2g (E)$\star$ 4g

The acceleration of gravity on the surface of any planet can be expressed in terms of its mass, radius and the gravitational constant G:

$$a=\frac{GM}{R^2}.$$

In case of the Earth, it is called g:

$$g=\frac{GM_E}{R_E^2}$$

At the surface of Planet X the acceleration due to gravity is four times larger than on the Earth:

$$a_X=\frac{GM_X}{R_X^2}=\frac{GM_E}{(R_E/2)^2}=4\frac{GM_E}{R^2}=4g.$$

6.

A satellite orbits the earth in an elliptical orbit, as shown in the figure. At which point in the figure does the satellite have the smallest speed?
(A) P (B) Same speed everywhere (C) Q (D) R (E)^* T

From the conservation of enery equation you can see that the speed is the smallest when r is the largest.

7.

Planet Vulcan has a mass of 10²³kg and a radius of 10⁶m. The crew of the spaceship Enterprise wishes to launch a 100kg space probe from Vulcan's surface. They ask you to calculate the total kinetic energy (in J) of the space probe at launch so that it can reach a maximum distance of 4.0$\times$10⁶m from Vulcan's center.
(A) 5$\times$ 10⁷ (B) 9$\times$ 10⁷ (C) 1$\times$ 10⁸ (D)$\star$ 5$\times$ 10⁸ (E) 8$\times$ 10⁸

The mechanical energy, which is the sum of the kinetic energy and potential energy is always conserved during the motion of the space probe: K₁+U₁=K₂+U₂, where the quantities with subscript 1 and 2 denote the energies at launch on the surface of Vulcan and at the distance desired to reach, respectively. The potential energies can be expressed as

$$U_1=-G\frac{Mm}{R}$$

and

$$U_2=-G\frac{Mm}{d}$$

where M and R are the mass and radius of Vulcan, m is the mass of the space probe and d is the maximum distance. At maximum distance, the speed and kinetic energy of the probe are zero: K₂ = 0, i.e.

$$K_1=U_2-U_1=GMm\left( \frac{1}{R}-\frac{1}{d}\right)=$$

$$=6.672\times10^{-11}\frac{Nm^2}{kg^2}\cdot 10^{23}kg\cdot 100kg \left(\frac{1}{10^6m}-\frac{1}{4\times 10^6m}\right)=$$

$$=\frac{6.672\times 10^{14}}{10^6}Nm\left(1-\frac{1}{4}\right)=\frac{3}{4}6.672\times10^8 J=5\times10^8J.$$

8.

A 1kg mass oscillates horizontally while attached to a spring. By what factor will the oscillation period change if we replace it by a 25kg mass?
(A) No change (B) 1/5 (C) 1/25 (D)$\star$ 5 (E) 25

The period of oscillation

$$T=2\pi\sqrt{\frac{m}{k}},$$

where k is the spring constant and m is the mass attached to the spring. You can see that if the mass is 25 times larger while the spring is the same, then the period of oscillation T would increase by a factor of $\sqrt{25}=5$.

9.

A 1kg mass suspended at the end of a light string oscillates back and forth in the vertical plane. By what factor will the oscillation period change if we replace it by a 25kg mass?
(A)$\star$ No change (B) 1/5 (C) 1/25 (D) 5 (E) 25

The period of the pendulum is expressed in terms of the length and the gravitational acceleration:

$$T=2\pi\sqrt{\frac{L}{g}},$$

i.e. it does not depend on the mass of the body suspended at the end of the string. Therefore, the period of oscillation is the same in both cases.

10.

A mass attached to a spring performs simple harmonic motion on a horizontal table with negligible friction. Consider the instant when its displacement from equilibrium is maximum. Which of the following quantities will decrease in magnitude at a later instant?
(A) Mechanical Energy (B)$\star$ Acceleration

(C) Period (D) Frequency

(E) Kinetic Energy

Mechanical Energy is conserved if friction is negligible.

The magnitude of acceleration is proportional to the displacement of the mass: $a=\frac{F}{m}=\frac{kx}{m}=\frac{k}{m} x$. At the point of maximum displacement the acceleration is maximum. When the mass starts to return back toward the equilibrium its displacement decreases and so does the acceleration.

The period and frequency does not change during the motion.

When the mass is at the position of maximum displacement it is at rest for an instant, i.e. its speed and kinetic energy is zero. As it starts to move again its speed and kinetic energy increase.

Problems

1.

I did a demonstration in class where I was sitting on a rotating chair holding a bicycle wheel whose rotation axis was pointing upward. In the beginning, the chair was at rest and the wheel was rotating counterclockwise as observed from above looking down on the axis of rotation.

a) ( 8 points) What happened when I flipped the wheel so that the axis of rotation was now pointing downward and the wheel was rotating clockwise as observed from above? Please provide a brief explanation of what caused the observed phenomenon.

The torque due to the external forces which act on the system of man-chair and rotating wheel is zero. Thus the total angular momentum of the system is conserved:

$$\Sigma \tau_{ext}=\frac{dL}{dt}=0.$$

The total angular momentum of the system has two components: the angular momentum of the wheel and the man-chair: L=L_w+L_m.

At the beginning, only the wheel was rotating (counterclockwise, so its angular momentum vector points upward - take this direction as positive) then the total angular momentum is

L_i=L_w.

When the wheel is flipped (rotates clockwise) the magnitude of its angular momentum is the same (if we neglect friction at the axis) but it points downward (negative): -L_w. In order to restore the original value of the total angular momentum the man-chair subsystem must start rotate counterclockwise producing a positive angular momentum in order to compensate the change in the angular momentum of the wheel:

$$L_f=-L_w+L_m=L_w=L_i \quad \rightarrow \quad L_m=2L_w\gt.$$

2.

In an in-class demonstration, I sat on a rotating chair holding two weights, each of mass 2kg. The moment of inertia of my body plus chair was approximately 6 kg m², not counting the moment of inertia of the two weights. When my arms were extended horizontally, the two weights were at a distance of 0.7m from the axis of rotation and I was rotating with an angular speed of 2.5 rad/s. I then pulled my arms in, which brought the two weights to a distance of 0.1m from the axis of rotation.

a) (7 points) Did my angular momentum increase or decrease after I pulled in the weights?

Since the torque due to the external forces is zero the angular momentum is conserved:

$$\Sigma \tau_{ext}=\frac{dL}{dt}=0 \quad \rightarrow \quad L=I\omega=constant$$

b) (9 points)

Calculate my angular speed after I pulled in the weights.

Since the angular momentum is conserved:

$$I_i \omega_i = I_f \omega_f,$$

the final angular speed is

$$\omega_f = \frac{I_i \omega_i}{I_f}.$$

The system consists of two parts: the man-chair and the 2 weights. The total moment of inertia is the sum of the subsystems' moment of inertia: I= I_m+2I_w, where I_w=mr² is the moment of inertia of one weight. The man-chair subsystem's moment of inertia does not change but the weights' does.

At the beginning:

$$I_i= 6kgm^2 + 2[ 2kg\cdot (0.7m)^2]=6kgm^2+1.96kgm^2=7.96kgm^2,$$

and at the end

$$I_f= 6kgm^2 + 2[ 2kg\cdot (0.1m)^2]=6kgm^2+0.04kgm^2=6.04kgm^2.$$

Then

$$\omega_f = \frac{7.96kgm^2 \cdot 2.5 rad/s}{6.04kgm^2}=3.3 rad/s.$$

c) (9 points) Calculate my kinetic energy before I pulled in the weights.

The kinetic energy is

$$K_i=\frac{1}{2}I_i\omega_i^2=\frac{1}{2} 7.96kgm^2 \cdot (2.5rad/s)^2=24.9 J$$

d) (7 points) After I pulled in the weights, did my kinetic energy increase or decrease? Explain briefly why. Can you account for the difference in kinetic energy before and after?

The kinetic energy increases:

$$K_f=\frac{1}{2}I_f\omega_f^2= \frac{1}{2} L \omega$$

where L is the angular momentum. Since the latter remains the same and since the angular speed must increase, the kinetic energy also increases.

The increase in the kinetic energy is due to the work done to move the weights closer to the rotation axis.