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AP Physics 1 : Circular, Rotational, and Harmonic Motion

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Example Questions

Example Question #41 : Harmonic Motion

Consider the following system:

Pendulum_1

If the length of the pendulum is and the maximum velocity of the block is $3\frac{m}{s}$ , what is the minimum possible value of angle A?

$g = 10\frac{m}{s^2}$

Possible Answers:

$51.4^{\circ}$

$50.8^{\circ}$

$67.7^{\circ}$

$56.9^{\circ}$

$74.2^{\circ}$

Correct answer:

$50.8^{\circ}$

Explanation:

We can use the equation for conservation of energy to solve this problem.

E = U_i+K_i = U_f+K_f

If the initial state is when the mass is at its highest position and the final state is when the mass is at its lowest position, then we can eliminate initial kinetic energy and final potential energy:

U_i = K_f

Substituting expressions in for each term, we get:

$mgh_i = \frac{1}{2}mv_f^2$

Canceling out mass and rearranging for height, we get:

$h_i = \frac{v_f^2}{2g}$

Thinking about a pendulum practically, we can write the height of the mass at any given point as a function of the length and angle of the pendulum:

h_i = l - lsin(A)

h_i = l(1-sin(A))

Think about how this formula is written. The second term gives us how far down the mass is from the top point. Therefore, we need to subtract this from the length of the pendulum to get how high above the lowest point (the height) the mass currently is.

Substituting this into the previous equation, we get:

$l(1-sin(A)) = \frac{v_f^2}{2g}$

Rearrange to solve for the angle:

$1-sin(A) = \frac{v_f^2}{2gl}$

$1-\frac{v_f^2}{2gl} = sin(A)$

$A = sin^{-1}(1-\frac{v_f^2}{2gl})$

We have values for each variable, allowing us to solve:

$A = sin^{-1}\left ( 1-\frac{(3\frac{m}{s})^2}{2(10\frac{m}{s^2})(2m)}\right ) = sin^{-1}(0.775)$

$A = 50.8^{\circ}$

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Example Question #22 : Pendulums

Matt Damon is once again trapped on Mars. He must measure the length of rope he has using only a stopwatch. Please solve the problem below.

A pendulum on Mars has been measured to have a period of 5.3 seconds. Using the knowledge that gravity on Mars is $3.711\frac{m}{s^2}$ determine the length of the simple pendulum. Round to 3 significant figures.

Possible Answers:

2.64 m

5.29m

8.30 m

2.64 cm

49.8 cm

Correct answer:

2.64 m

Explanation:

To find the answer one must manipulate the equation

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

Where represents the period of the motion, the length of the pendulum, and the gravity or acceleration the system is under.

To solve this for we will start by dividing both sides by $2\pi$ . Next will with square both sides and finally multiply by , to come to the form below

$\frac{gT^{2}}{4\pi^{2}} = L$

Now plugging in our numbers

$\frac{3.711\frac{m}{s^2}*(5.3s)^{2}}{4*3.14^{2}} = \frac{104.24199}{39.4384} = 2.6431597 = 2.64$

Keep in mind that the most accurate method is to round numbers at the very end of calculations (above this isn't done from the start of pi for simplicity).

The unit calculation above will end with meters as we are taking $\frac{m}{s^{2}}*s^{2}$ which will leave as the final unit of your answer.

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Example Question #1 : Other Harmonic Systems

You push on a door with a force of 1.3 N at an angle of 45 degrees to the surface and at a distance of 0.5 m from the hinges. What is the torque produced?

Possible Answers:

$0.46N\cdot m$

$0.64N\cdot m$

$0.31N\cdot m$

$0.84N\cdot m$

$1.2N\cdot m$

Correct answer:

$0.46N\cdot m$

Explanation:

To calculate torque, this equation is needed:

$\tau=rFsin\Theta$

Next, identify the given information:

r= 0.5 m

F=1.3N

$\Theta=45^{o}$

Plug these numbers into the equation to determine the torque:

$\tau=rFsin\Theta = (0.5m)(1.5N)sin(45^{o})=0.46N\cdot m$

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Example Question #43 : Circular, Rotational, And Harmonic Motion

Photo_17

If a mass on the end of a string of length 8 cm is pulled 30 degrees away from vertical, what will its speed be when the string is vertically alligned if the mass is released from rest?

$g=9.81\frac{m}{s^2}$

Possible Answers:

$2.1\frac{m}{s}$

$1.3\frac{m}{s}$

$4.5\frac{m}{s}$

$1.7\frac{m}{s}$

$0.46\frac{m}{s}$

Correct answer:

$0.46\frac{m}{s}$

Explanation:

We can determine the vertical components by using some geometry.

Photo_2

Since this problem boils down to being a problem of conservation of energy, we can state that

$E_{i}=E_{f}\rightarrow PE_{i}+KE_{i}=PE_{f}+KE_{f}$

Since the mass is released from rest and we can state at the bottom of its arc it will be at height 0 m, this equation can be simplified:

$PE_{i}=KE_{f} \rightarrow mgh=\frac{1}{2}mv^{2} \rightarrow gh=\frac{1}{2}v^{2}$

is equal to the height of the mass above 0 m; therefore, by referring to the diagram, we can determine that $h=R-Rcos\Theta$ . By plugging this in and rearranging the equation, we can solve for the speed of the mass when the string is vertically aligned:

$v=\sqrt{2gR(1-cos\Theta)}=\sqrt{2(9.81\frac{m}{s^2})(8\cdot10^{-2}m)(1-cos30^{o})}=0.46\frac{m}{s}$

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Example Question #41 : Harmonic Motion

A ball with mass 5 kg is attached to a spring and is released 10 meters from equilibrium. After some time, the ball passes the equilibrium point moving at $7\frac{m}{s}$ . What is the spring constant () of the spring?

Possible Answers:

$2.45 \frac{kg}{s^2}$

$1.225\frac{kg}{s^2}$

$0.41\frac{kg}{s^2}$

$3.5\frac{kg}{s^2}$

$0.82\frac{kg}{s^2}$

Correct answer:

$2.45 \frac{kg}{s^2}$

Explanation:

When the object is released, it has no kinetic energy (it isn't moving) and a potential energy of

$P.E.=\frac{1}{2}kx^2=\frac{1}{2}k(10m)^2=50m^2\cdot k$

When the object passes through the equilibrium point, it has no potential energy ( x=0 ) and a kinetic energy of

$K.E.=\frac{1}{2}mv^2=\frac{1}{2}(5kg)(7\frac{m}{s})^2=122.5\frac{kg\cdot m^2}{s^2}$

Due to the conservation of energy, these two quantities must be equal to each other:

$50m^2\cdot k=122.5\frac{kg*m^2}{s^2}$

$k = \frac{122.5\frac{kg\cdot m^2}{s^2}}{50m^2}=2.45\frac{kg}{s^2}$

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Example Question #1 : Other Harmonic Systems

A 30 kg weight attached to a spring is at equilibrium lying horizontally on a table. The spring is lifted up and is stretched by 80 cm before the weight is lifted off the table. What is this spring's spring constant ()?

Possible Answers:

$0.375\frac{kg}{s^2}$

$460\frac{kg}{s^2}$

$37.5\frac{kg}{s^2}$

$3.68\frac{kg}{s^2}$

$368\frac{kg}{s^2}$

Correct answer:

$368\frac{kg}{s^2}$

Explanation:

First, we should convert 80 centimeters to 0.8 meters.

We know the force applied to the weight by gravity is

F_g=mg

$294N=(30kg)(9.81\frac{m}{s^2})$

The force applied by the spring in the opposite direction must be equal to this:

-kx=-294N

k(0.8m) = 294N

$k=368\frac{N}{m}=368\frac{kg}{s^2}$

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Example Question #1 : Other Harmonic Systems

Instead of using a second hand to count seconds, a watchmaker decides to construct a simple harmonic system involving a mass and a spring suspended from a table.

Ignoring the effects of gravity, if the spring the watchmaker selects has a spring constant of $0.01\frac{kg}{s^2}$ , how large of a mass in kilograms should he attach to the spring such that the harmonic system does not oscillate too slow or too fast?

Possible Answers:

$\frac{1}{3}kg$

$\frac{3}{\pi^2}kg$

$\frac{9}{10}kg$

$\frac{9}{\pi^2}kg$

$\frac{1}{6}kg$

Correct answer:

$\frac{9}{\pi^2}kg$

Explanation:

The angular velocity of the harmonic system is equal to the square root of the spring constant over the mass:

$\omega=\sqrt{\frac{k}{m}}$

Since we need the angular velocity to be $\frac{2\pi}{60}$ and we are given the spring constant as $0.01\frac{kg}{s^2}$ , then we can set up an equation:

$\frac{2\pi}{60}=\sqrt{\frac{0.01}{m}}$

$\frac{4\pi^2}{60^2}=\frac{0.01}{m}$

$m=\frac{60^2*0.01}{4\pi^2}=\frac{9}{\pi^2}kg$

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Example Question #5 : Other Harmonic Systems

The position of a 1kg mass in an oscillating spring-mass system is given by the following equation:

$x(t)=2sin(4\pi t)+4$ , where is measured in seconds , and is measured in meters .

What is the velocity equation of the system as a function of time?

Possible Answers:

$v(t)=8\pi cos( t)$

$v(t)=2 cos(4\pi t)$

$v(t)=2 sin(4\pi t)$

$v(t)=8\pi sin(4\pi t)$

$v(t)=8\pi cos(4\pi t)$

Correct answer:

$v(t)=8\pi cos(4\pi t)$

Explanation:

The velocity equation can be found by differentiating the position equation.

$v(t)=x'(t)=\frac{d}{dt}(2sin(4\pi t)+4)=2*4\pi cos(4\pi t)$

$=8\pi cos(4\pi t)$

Here, we made use of the chain rule in taking the derivative.

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Example Question #1 : Circular And Rotational Motion

A horizontally mounted wheel of radius is initially at rest, and then begins to accelerate constantly until it has reached an angular velocity $\omega$ after 5 complete revolutions. What was the angular acceleration of the wheel?

Possible Answers:

$\frac{\omega}{10\pi }$

$\frac{\omega ^{2}}{5\pi r}$

$\frac{\omega ^{2}}{20\pi }$

$\frac{\omega ^{2}}{10\pi}$

$\frac{\omega ^{2}}{10\pi r}$

Correct answer:

$\frac{\omega ^{2}}{20\pi }$

Explanation:

You may recall the kinematic equation that relates final velocity, initial velocity, acceleration, and distance, respectively:

$v_{f}^{2}=v_{i}^{2}+2ad$

Well, for rotational motion (such as in this problem), there is a similar equation, except it relates final angular velocity, intial angular velocity, angular acceleration, and angular distance, respectively:

$\omega _{f}^{2}=\omega _{i}^{2}+2\alpha \theta$

The wheel starts at rest, so the initial angular velocity, $\omega _{i}$ , is zero. The total number of revolutions of the wheel is given to be 5 revolutions. Each revolution is equivalent to an angular distance of $2\pi$ radians. So, we can convert the total revolutions to an angular distance to get:

$\theta=5\:rev\cdot \frac{2\pi }{1\:rev}=10\pi$

The final angular velocity was given as $\omega$ in the text of the question. So, we should use the above equation to solve for the angular acceleration, $\alpha$ .

$\omega ^{2}=0+2\alpha\cdot 10\pi$

$\alpha=\frac{\omega ^{2}}{20\pi }$

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Example Question #1 : Angular Velocity And Acceleration

An object moves at a constant speed of $9.0 \:\frac{m}{s}$ in a circular path of radius of 1.5 m. What is the angular acceleration of the object?

Possible Answers:

$36\:\frac{rad}{s^2}$

$81\:\frac{rad}{s^2}$

$54\:\frac{rad}{s^2}$

$14\:\frac{rad}{s^2}$

$6.0\:\frac{rad}{s^2}$

Correct answer:

$36\:\frac{rad}{s^2}$

Explanation:

For a rotating object, or an object moving in a circular path, the relationship between angular acceleration and linear acceleration is

$a=\alpha r$

Linear acceleration is given by , angular acceleration is $\alpha$ , and the radius of the circular path is .

For circular/centripetal motion, the linear acceleration is related to the object's linear velocity by

$a_{centripetal}=\frac{v^{2}}{r}$

We know the linear velocity is $9.0 \:\frac{m}{s}$ , and the radius is 1.5 m, so we can find the linear acceleration...

$a_{c}=\frac{(9.0\:\frac{m}{s})^{2}}{1.5\:m}=54\:\frac{m}{s^2}$

Now that we have the linear acceleration, we can use this in the equation at the top to find the angular acceleration...

$\alpha=\frac{a}{r}=\frac{54\:\frac{m}{s^2}}{1.5\Lm}=36\:\frac{rad}{s^2}$

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AP Physics 1 : Circular, Rotational, and Harmonic Motion

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #41 : Harmonic Motion

Example Question #22 : Pendulums

Example Question #1 : Other Harmonic Systems

Example Question #43 : Circular, Rotational, And Harmonic Motion

Example Question #41 : Harmonic Motion

Example Question #1 : Other Harmonic Systems

Example Question #1 : Other Harmonic Systems

Example Question #5 : Other Harmonic Systems

Example Question #1 : Circular And Rotational Motion

Example Question #1 : Angular Velocity And Acceleration

All AP Physics 1 Resources

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