Rotational, Circular, and Harmonic Motion - MCAT Physical

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Question

Which factors increase the maximum velocity of a pendulum?

Answer

Both the length of the pendulum's string and the angle of displacement affect the maximum velocity of the pendulum. Increasing the length of the pendulum's string and increasing the angle of displacement both increase the distance the pendulum must travel in a single period, increasing its potential energy at its maximum height, and therefore the maximum velocity at its lowest point.

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Question

What is the period of a pendulum that has a string length of 9.8m?

Answer

The key to answering this question is to recall the following important formula for a simple pendulum: $T = 2\pi \sqrt{\frac{L}{g}}$ .

$T=2\pi\sqrt{\frac{9.8}{9.8}}=2\pi$

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Question

Which of the following changes to a pendulum will affect the angular velocity?

Answer

For a pendulum, the angular velocity is given by the equation $\omega = \sqrt{\frac{g}{L}}$ , where is the acceleration due to gravity and is the length of the pendulum. Of the available answer choices, only changing the length of the string will effect the angular velocity. $\omega$ does not depend on mass or the release point of the pendulum.

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Question

An insect sits at the edge of slowly turning wheel. The wheel is accelerated gradually until the insect can no longer hold on. The insect’s path of travel will be:

The insect will move away from the wheel in a straight line perpendicular to the wheel edge.

The insect will move away from the wheel in a straight line tangent to the wheel edge.

The insect will move away from the wheel edge in a curvilinear path.

The path is not predictable by Newtonian physics.

None of these is true.

Answer

Choice 2 is correct; this is how a classical slingshot operates. A body in motion tends to remain in motion unless a force disturbs that motion. At any point in time, the insect is moving in a straight line along a line tangent to the circumference of the wheel. When the little creature leaves the wheel, there is no longer any force acting on it, so it will move in a straight line, not a curvilinear path. Newtonian kinetics of course includes these concepts, because they relate closely to the motion of celestial bodies.

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Question

For an object traveling in a circle at a constant velocity of 100m/s, which of the following is true?

Answer

Since the object is traveling in a circle, it is constantly changing direction. This means that the direction of the velocity is also changing, even if the magnitude is not. Because the magnitude of the velocity does not change, we can assume that acceleration is constant.

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Question

A 2kg mass is suspended on a rope that wraps around a frictionless pulley. The pulley is attached to the ceiling and has a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.

What is the linear velocity of the pulley the instant the rope to the 2kg mass is cut?

Answer

Note that we are asked the linear velocity of the pulley, not the angular velocity. With Newton’s second law, we know that F = ma. Additionally, in rotational motion, we know that $F=\frac{mv^2}{r}$ . The force (F) the instant the rope is cut is due to the platform with the 4 weights. Knowing the mass and radius of the pulley, we can solve for the linear velocity:

$F=\frac{mv^2}{r}$

$v=\sqrt{\frac{Fr}{m}}=\sqrt{\frac{(19.6N)(0.25m)}{0.01kg}}=22.1m/s$

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Question

A ball with a mass of 0.95kg is attached to a string with a length of 75cm and moves in a vertical circle at a constant speed of 25m/s. What is the maximum tension in the string?

Answer

Maximum tension occurs when the ball is at its bottommost point, as the string experiences both the downward force of gravity and centripetal force.

$F_{c} = \frac{mv^{2}}{r}$

$F_{g} = mg$

$F_{net} = F_{c } + F_{g}$

$F_{net} = \frac{mv^{2}}{r} + mg = \frac{(0.95 kg)(25 m/s)^{2}}{0.75 m} + (0.95 kg)(10 m/s^{2}) = 800 N$

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Question

Two identical cars, A and B, are traveling on identical surfaces. Car A drives at a constant velocity in a perfectly circular path. Car B drives at a constant velocity along a long, straight path. Which best describes the accelerations, $\small a_{A}$ and $\small a_{B}$ , of the two cars?

Answer

An object moving in constant circular motion must have centripetal acceleration, directed towards the center of its path, in order to maintain its circular motion. So $\small a_{A} eq0$ and is towards the center of car A's path.

An object moving linearly at constant velocity has 0 acceleration, so $\small a_{B}=0$ .

In general, we can think of acceleration as a change in either the magnitude or direction of an object's velocity. Neither of these cars has a change in the magnitude of its velocity, but car A does experience a change in the direction of its velocity, while car B does not.

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Question

What is the approximate centripetal acceleration of a point on the rim of a bicycle wheel with a diameter of $\small 60cm$ if the wheel is rotating at eight revolutions per second?

Answer

Centripetal acceleration is given by the equation:

$a_c=\frac{v^2}{r}$

We are given the angular velocity and the diameter. We can easily solve for the radius from the diameter, and use the angular velocity to find the linear velocity.

Convert the angular velocity to the proper units by changing revolutions to radians.

$\omega=\frac{8rev}{s}\frac{2\pi rad}{1rev}=16\pi\frac{rad}{s}$

Now we can convert the angular velocity to the linear velocity using the radius.

$v=\omegar=16\pi\frac{rad}{s}*0.30m=4.8\pi\frac{m}{s}$

Use the radius and the linear velocity to calculate the centripetal acceleration from the original equation.

$a_c=\frac{v^2}{r}=\frac{(4.8\pi\frac{m}{s})^2}{0.30m}$

$a_c=\frac{23.04\pi^2\frac{m^2}{s^2}}{0.30m}$

$a_c=76.8\pi^2\frac{m}{s^2}$

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Question

What is the centripetal force on a $\small 10kg$ ball being swung in a vertical loop with a velocity of $\small 5\frac{m}{s}$ on a $\small 1m$ string?

Answer

Centripetal force is given by the equation:

$F_c = ma_c=m\frac{v^2}{r}$

The length of the string represents the radius of the circle being formed. Use the given string length as the radius, along with the mass of the ball and the velocity, to calculate the centripetal force.

$F_c=(10kg)\frac{(5\frac{m}{s})^2}{1m}$

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Question

A man of mass is riding a roller coaster, which has a total mass of . As the ride enters a circular loop, it is traveling at a rate of $20 \frac{m}{s}$ . The loop has a radius of . If the average frictional force on the coaster is , how many g's of force does the man experience at the top of the loop?

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

Answer

To calculate g's, we need to know the centripetal force the rider is experiencing. To calculate centripetal force, we need to know how fast the ride is traveling at the top of the loop. To do this, we will use the equation for conservation of energy:

If we set the height of the top of the loop to 10m, we can say that there is no initial potential energy. Since we are solving for final velocity, let's rearrange this for final kinetic energy:

Plugging in the expressions for each of these terms, we get:

$\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2-mgh_f - F_fd$

Rearranging for final velocity:

$v_f = \sqrt{v_i^2-2gh_f-\frac{2F_fd}{m}}$

We have all of the values except for how far the coaster travels. It travels from the bottom to the top of the loop, which is half the circumference of the circle:

$d = \pi r = 5\pi$

Now we can plug in our values and solve:

$v_f = \sqrt{(20\frac{m}{s})^2-(2\cdot10\frac{m}{s^2}\cdot10m)-\frac{2\cdot600N\cdot5\pi}{1075kg}}$

$v_f = \sqrt{400\frac{m^2}{s^2} - 200\frac{m^2}{s^2} - 5.58\pi\frac{m^2}{s^2}} = \sqrt{182.47\frac{m^2}{s^2}} = 13.51 \frac{m}{s}$

Now that we know how fast the ride is traveling at the top of the loop, we can calculate the centripetal force on the man:

$F_c = ma_c = m\frac{v^2}{r}$

$F_c = (75kg)\frac{(13.51\frac{m}{s})^2}{5m} = 2737 N$

Now that we know the force in Newtons, we can convert this value to g's.

For this man:

$1'g' = mg = (75kg)(10\frac{m}{s^2}) = 750 N$

Therefore,

$G = \frac{2737N}{750\frac{N}{g}} = 3.65g$

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Question

A spring is compressed as far possible and is not permitted to expand. What can be said about its potential energy and its kinetic energy?

Answer

In this case, the kinetic energy of the spring is at a minimum. This is because, as the question indicates, the spring is not moving. At the same time, because the spring is compressed as far is it can be compressed, we know that its potential energy is at a maximum. The total energy of the spring therefore cannot be zero.

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Question

Which of the following changes to a pendulum would decrease its period?

Answer

The only factor that affects the period of a pendulum is the length of the pendulum. Therefore we can ignore any of the other answers which include other factors. The equation for the period of a pendulum is:

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

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Question

A long pendulum with a length of has a mass attached to the end of it. Approximately how long does it take for the pendulum to swing one time from its maximum displacement?

Answer

We are asked for the period of the pendulum, or the time it takes to make one full revolution or swing.

The period of a pendulum is given by this formula:

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

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

$T=2\pi\sqrt{\frac{40}{10}}=2\pi\sqrt{\4}=T=2\pi\cdot2=12s$

It takes roughly 12 seconds for the pendulum to make a full swing. Note that the mass attached to the end of the pendulum is irrelevant.

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Question

A child of mass m stands at the outer edge of a carousel of radius R, which is rotating with angular velocity $\small \omega$ . If the child then moves inwards, to a distance of R/4 from the center of the carousel, which of the following statements regarding the child-carousel system is true?

Answer

Due to conservation of angular momentum, the angular momentum L must remain constant: $\small \small L_{i}=L_{f}$ .

Because of the child’s decreased distance from the axis of rotation, the moment of inertia of the system also decreases The child can be approximated as a point mass with moment of inertia $\small \small I = mr^{2}$ , and the carousel’s moment of inertia remains unchanged. Moving the child toward the center of the carousel will decrease its moment of inertia.

To examine what happens to angular velocity, we need the definition of angular momentum, $\small L = I\omega$ . Replacing this in the conservation of angular momentum equation, we see that $\small \small I_{i}\omega_{i}=I_{f}\omega_{f}$ . Since $\small \small I_{f}<I_{i}$ , then $\small \small \omega_{f}>\omega_{i}$ so that the product remains the same.

Angular momentum remains the same, moment of inertia decreases, and angular velocity increases.

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Question

A carousel is rotating clockwise when observed from a bird's-eye view (looking directly down from above). What is the direction of the angular velocity vector, $\small \omega$ ?

Answer

We can apply the right-hand rule for angular velocity: if the fingers on the right hand curl in the direction of rotation, the thumb points in the direction of the angular velocity vector. In this case the fingers curl clockwise, so thumb points downwards.

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Question

A 2kg mass is suspended on a rope that wraps around a frictionless pulley attached to the ceiling with a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.

If we assumed the pulley had internal friction, how would the angular acceleration of the pulley change?

Answer

The angular acceleration would decrease. Newton’s second law states that F = ma. The net force, however, is reduced in the internal friction scenario because the friction acts to oppose the motion created by the rope around the pulley. As a side note, the pulley would also heat up because the energy change due to friction is released as heat.

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Question

A solid sphere, a hollow sphere, and a thin solid disk each have mass M and radius R and are released from rest at the top of a frictionless inclined plane. In which order do the objects reach the end of the inclined plane?

Assume that all objects roll, rather than slide.

Answer

An object with low moment of inertia will acquire less rotational kinetic energy and more translational kinetic energy (moves faster down the ramp) than an object with higher moment of inertia.

Moment of inertia depends on how close to the object's center of mass its mass is concentrated. An object with more of its mass close to the center will have lower moment of inertia, and subsequently more translational kinetic energy. The solid sphere has lowest moment of inertia, then the disk, and then the hollow sphere, thus the solid sphere reaches the end first, then the disk, and finally the hollow sphere.

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Question

An eight inch long cylinder is composed of two substances: the first four inches are composed of aluminum, and the second four inches are composed of lead. The aluminum region has a mass of 30g and the lead region has a mass of 60g.

Where is the center of mass in the cylinder?

Answer

The center of mass is the point where we can consider all of the substance's mass to be concentrated. Finding the center of mass greatly facilitates mechanics problems, because we can perceive the force to be acting on only this one point.

Since the lead side of the cylinder is heavier than the aluminum side, we can conclude that the center of mass is located inside the lead side of the cylinder. Since the lead side has a mass of 60g, each inch of the lead side will have a mass of 15g. By going one inch into the lead side, there are 45g on each side of the point. As a result, we say that this is the location for the center of mass for the cylinder, as either side would weigh the same individually.

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Question

Scientists studying the collision of particles note that a certain particle travels around a collider two times every second. If the tube's radius is , with what velocity is the particle traveling?

Answer

The equation relating velocity and radius is $\omega = \frac{ u }{r}$ .

Because angular velocity can also be found using the equation $\omega = 2\pi f$ , we can set the equations equal and solve for the velocity.

$\frac{v}{r}=2\pi f$

$v=2\pi f r$

Since we are given the radius as and the frequency as , we can calculate the velocity. (Frequency can be found by the number of times an object travels around a full circle in one second).

$u = 2 \pi (2Hz)(5000m) = 2\pi * 10^{4}\frac{m}{s}$

Alternatively, we can use the radius to determine the circumference of the collider.

$C=2\pi r=2\pi (5000)=10000\pi$

The particle travels twice around the collider every second.

$v=\frac{d}{t}=\frac{2(10000\pi)m}{1s}=2\pi*10^4\frac{m}{s}$

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