Circular, Rotational, and Harmonic Motion - AP Physics 1

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Question

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?

Answer

To calculate torque, this equation is needed:

$\tau=rFsin\Theta$

Next, identify the given information:

$\Theta=45^{o}$

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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Question

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?

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

Answer

We can determine the vertical components by using some geometry.

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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Question

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?

Answer

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 () 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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Question

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 ()?

Answer

First, we should convert 80 centimeters to 0.8 meters.

We know the force applied to the weight by gravity is

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

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

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

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Question

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?

Answer

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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Question

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

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

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

Answer

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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Question

A ball of mass 2kg is attached to a string of length 4m, forming a pendulum. If the string is raised to have an angle of 30 degrees below the horizontal and released, what is the velocity of the ball as it passes through its lowest point?

Answer

This question deals with conservation of energy in the form of a pendulum. The equation for conservation of energy is:

According to the problem statement, there is no initial kinetic energy and no final potential energy. The equation becomes:

Subsituting in the expressions for potential and kinetic energy, we get:

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

We can eliminate mass to get:

$gh = \frac{1}{2}v_f^2$

Rearranging for final velocity, we get:

$v_f = \sqrt{2gh}$

In order to solve for the velocity, we need to find the initial height of the ball.

The following diagram will help visualize the system:

From this, we can write:

Using the length of string and the angle it's held at, we can solve for :

$d = 4sin(30^{\circ}) = 2$

Now that we have all of our information, we can solve for the final velocity:

$v_f = \sqrt{2\cdot 10\cdot 2} = \sqrt{40} = 6.3 \frac{m}{s}$

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Question

A pendulum has a period of 5 seconds. If the length of the string of the pendulum is quadrupled, what is the new period of the pendulum?

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

Answer

We need to know how to calculate the period of a pendulum to solve this problem. The formula for period is:

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

In the problem, we are only changing the length of the string. Therefore, we can rewrite the equation for each scenario:

$P_1 = \frac{2\pi}{\sqrt{g}} \sqrt {{L_1}}$

$P_2 = \frac{2\pi}{\sqrt{g}} \sqrt {{L_2}}$

Dividng one expression by the other, we get a ratio:

$\frac{P_1}{P_2} = \sqrt{\frac{L_1}{L_2}}$

We know that , so we can rewrite the expression as:

$\frac{P_1}{P_2} = \sqrt{\frac{L_1}{4L_1}} = \sqrt{\frac{1}{4}}=\frac{1}{2}$

Rearranging for P2, we get:

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Question

A student studying Newtonian mechanics in the 19th century was skeptical of some of Newton's concepts. The student has a pendulum that has a period of 3 seconds while sitting on his desk. He attaches the pendulum to a ballon and drops it off the roof of a university building, which is 20m tall. Another student realizes that the pendulum strikes the ground with a velocity of $12\frac{m}{s}$ . What is the period of the pendulum as it is falling to the ground?

Neglect air resistance and assume $g=10\frac{m}{s^2}$

Answer

We need to know the formula for the period of a pendulum to solve this problem:

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

We aren't given the length of the pendulum, but that's ok. We could solve for it, but it would be an unnecessary step since the length remains constant.

We can write this formula for the pendulum when it is on the student's table and when it is falling:

$T_1 = 2\pi\sqrt{\frac{L}{g_1}}$

$T_2 = 2\pi\sqrt{\frac{L}{g_2}}$

1 denotes on the table and 2 denotes falling. The only thing that is different between the two states is the period and the gravity (technically the acceleration of the whole system, but this is the form in which you are most likely to see the formula). We can divide the two expressions to get a ratio:

$\frac{T_2}{T_1} = \frac{2\pi\sqrt{\frac{L}{g_2}}}{2\pi\sqrt\frac{L}{g_1}}$

Canceling out the constants and rearranging, we get:

$\frac{T_2}{T_1} = \sqrt\frac{g_1}{g_2}$

We know g1; it's simply 10. However, we need to calculate g2, which is the rate at which the pendulum and balloon are accelerating toward the ground. We are given enough information to use the following formula to determine this:

$v_f^2 = v_i^2 + 2a\Delta x$

Removing initial velocity and rearranging for acceleration, we get:

$a = \frac{v_f^2}{2\Delta x}$

Plugging in our values:

$a = \frac{144\frac{m^2}{s^2}}{2\cdot20m} = 3.6 \frac{m}{s^2}$

This is our g2. We now have all of the values to solve for T2:

$T_2 = T_1 \sqrt{\frac{g_1}{g_2}}= (3s)\sqrt{\frac{10\frac{m}{s^2}}{3.6\frac{m}{s^2}}}$

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Question

Consider the following system:

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?

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

Answer

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

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:

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:

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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Question

A pendulum of mass has a period . If the mass is quadrupled to , what is the new period of the pendulum in terms of ?

Answer

The mass of a pendulum has no effect on its period. The equation for the period of a pendulum is

$T = 2\pi \sqrt{{\frac{L}{g}}}$ , which does not include mass.

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Question

In the lab, a student has created a pendulum by hanging a weight from a string. The student releases the pendulum from rest and uses a sensor and computer to find the equation of motion for the pendulum: $x(t)=0.13m;cos(6.3\frac{radians}{s}t+0.14:radians)$

The student then replaces the weight with a weight whose mass, is twice as large as that of the original weight without changing the length of the string. The student again releases the weight from rest from the same displacement from equilibrium. What would the new equation of motion be for the pendulum?

Answer

The period and frequency of a pendulum depend only on its length and the gravity force constant, . Changing the mass of the pendulum does not affect the frequency, and since the student released the new pendulum from the same displacement as the old, the amplitude and phase remain the same, and the equation of motion is the same for both pendula.

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Question

In the lab, a student has created a pendulum by hanging a weight from a string. The student releases the pendulum from rest and uses a sensor and computer to find the equation of motion for the pendulum: $x(t)=0.13m;cos(6.3\frac{radians}{s}t+0.14:radians)$

The student then replaces the string with a string whose length, $\l$ is twice as large as that of the original string without changing the mass of the weight. The student again releases the weight from rest from the same displacement from equilibrium. What would the new equation of motion be for the pendulum?

Answer

Doubling the length of a pendulum increases the period, so it decreases the frequency of the pendulum. The frequency depends upon the square root of the length, so the frequency decreases by a factor of $\sqrt{2}=1.41$ . Neither of the other parameters (amplitude, phase) change.

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Question

A pendulum of length will take how long to complete one period of its swing?

Answer

The period of a pendulum is given by the following formula:

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

Substituting our values, we obtain:

$T= \frac{2\pi }{\sqrt{\frac{10}{10}}} = \frac{2\pi}{1}\approx6.3$

Roughly 6.3 seconds is the time it takes for the pendulum to complete one period.

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Question

If a simple pendulum is set to oscillate on Earth, it has a period of . Now suppose this same pendulum were moved to the Moon, where the gravitational field is 6 times less than that of Earth.

What is the period of this pendulum on the Moon in terms of ?

Answer

The period of a simple pendulum is given by:

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

Where is the period of the pendulum, is the length of the pendulum, and is the gravitational constant of the planet we are on. Thus on Earth, the period is given by: $T_e=2\pi\sqrt{\frac{L}{g_e}}$

With being Earth's gravitational constant. The period on the Moon is given by:

$T_m=2\pi\sqrt{\frac{L}{g_m}}$

With being the Moon's gravitational constant. Since the Moon's gravity is 6 times weaker than that of Earth's, we have:

$g_m=\frac{g_e}{6}$

Plug this value into the Moon pendulum equation:

$T_m=2\pi\sqrt{\frac{L}{\frac{g_e}{6}}}=2\pi\sqrt{\frac{6L}{g}}=2\pi\sqrt{\frac{L}{g_e}}\sqrt{6}$

Since $T_e=2\pi\sqrt{\frac{L}{g_e}}$ ,

Substituting this into the above expression gives us $T_m=T_e\sqrt{6}$

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Question

A pendulum of length has a mass of attached to the bottom. Determine the frequency of the pendulum if it is released from a shallow angle.

Answer

The frequency of a pendulum is given by:

$\omega=\sqrt{\frac{g}{L}}$

Where is the length of the pendulum and is the gravity constant. Notice how the frequency is independent of mass.

Plugging in values:

$\omega=\sqrt{\frac{9.8}{2.5}}$

$\omega=\frac{1.98}{s}$

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Question

How will increasing the mass at the end of a pendulum change the period of it's motion? Assume a shallow angle of release.

Answer

The frequency of a pendulum is given by:

$\omega=\sqrt{\frac{g}{L}}$

Where is the length of the pendulum and is the gravity constant. The frequency is independent of mass. Thus, adding mass will have no effect.

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Question

Which simple pendulum will have a longer period?

$A:m = 10\textup{ kg}, L = 2\textup{ m}$

$B: m = 4\textup{ kg}, L = 3\textup{ m}$

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

Answer

The expression for the period of a pendulum is:

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

Therefore, the period of a pendulum is proportional to the square root of the length of the pendulum (assuming they are both on earth, or the same planetary body). Thus, the pendulum with the longer length will have the longer period.

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Question

A simple pendulum of length $L = 0.6\textup{ m}$ with a block of mass $m = 5\textup{ kg}$ attached has a maximum velocity of $1.2\ \frac{\textup{m}}{\textup{s}^2}$ . What is the maximum height of the block?

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

Answer

This may come as a surprise, but we don't need to know a single formula concerning pendulums or circular motion to solve this problem. We just have to be able to understand the motion of a pendulum and think about the situation practically. A pendulum reaches its maximum velocity when the block is at its lowest point (the pendulum is vertical and pointing straight down). We can then use the expression for conservation of energy to determine the maximum height of the block.

If we assume that the low point of the pendulum has a height of 0, we can eliminate initial potential energy. We can also eliminate final kinetic energy (at least for now. If we get a maximum height that is more than twice the length of the pendulum, we know that it completes full rotations) since the block will be at rest when it reaches its maximum height.

Substituting expressions in for each term:

$\frac{1}{2}mv_i^2=mgh_f$

Eliminating mass and rearranging for final height, we get:

$h_f = \frac{v^2}{2g}$

Plugging in our values, we get:

$h_f = \frac{\left ( 1.2\frac{m}{s}\right )^2}{2\left ( 10\frac{m}{s^2}\right )}$

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Question

A simple pendulum with a length of $L = 2\textup{ m}$ has a block of mass $m = 6\textup{ kg}$ attached to the end. If the pendulum is $h=1\textup{ m}$ above its lowest point and rotating downward, what is the instantaneous acceleration of the block?

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

Answer

Since we are told that the block is 1m above its lowest point (half way between its low point and horizontal), we can calculate the angle that the pendulum makes with the vertical:

$sin(\theta) = \frac{h}{L}= \frac{1m}{2m} = \frac{1}{2}$

$\theta = 30^{\circ}$

Then we can use the following expression to determine the net force acting on the block in the direction of its motion:

$F_{net}= mgsin(\theta)$

Then we can use Newton's 2nd law to determine the instantaneous acceleration:

$F_{net}=ma = mgsin(\theta)$

Canceling out mass and rearranging for acceleration:

$a = gsin(\theta)$

$a = \left ( 10\frac{m}{s^2}\right )\left ( \frac{1}{2}\right )$

$a = 5\frac{m}{s^2}$

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