Centripetal Force and Acceleration

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AP Physics 1 › Centripetal Force and Acceleration

Questions 1 - 10
1

A object moves in a circular motion with a diameter of . What is the magnitude of this object's velocity?

Explanation

Let's start with what we know, and what we want to know. We are given the mass of an object, and we're told that it's traveling in a circular orbit. Also, we have been given the diameter of the orbit. What we are trying to find is the velocity of the object.

In this problem, we'll need to consider the centripetal force of the object as it travels in a circular orbit. The force that is contributing to the centripetal force is the weight of the object. Therefore, we can write the following expression:

The above expression tells us that the object's velocity is dependent on the acceleration due to gravity, as well as the radius of the orbit. Since we're told that the diameter of the orbit is , we can find the radius by taking half of this value, which is . Moreover, the plus-or-minus sign indicates that the object can either be traveling in a clockwise or counter-clockwise orbit. In either case, the object will have the same magnitude of velocity.

Solving for velocity, we can plug in values to obtain:

2

A child is swinging a toy rocket around their head with a rope. If they double the frequency of "orbits", how will the tension in the rope change?

Quadrupled

Doubled

Halved

Quartered

None of these

Explanation

The tension in the rope is equal to the centripetal force on the rocket.

Definition of velocity:

The time necessary to go around one circle is it's period, which is the inverse of the frequency.

Doubling the frequency would double the velocity, which would quadruple the tension.

3

Pluto distance to sun:

Determine the centripetal acceleration of Pluto with respect to the Sun.

Explanation

Combine equations:

Convert to meters and seconds and plug in values:

4

Consider the following system:

Spinning rod with masses at end

Two spherical masses, A and B, are attached to the end of a rigid rod with length l. The rod is attached to a fixed point, p, which is at a height, h, above the ground. The rod spins around the fixed point in a vertical circle that is traced in grey. is the angle at which the rod makes with the horizontal at any given time ( in the figure).

If the maximum centripetal force felt by mass A is , what is the minimum centripetal force it feels? Neglect air resistance and internal friction.

Explanation

First we need to determine when at which points mass A will be experiencing the most/least centripetal force. The most will be when the rod is vertical and mass A is at the bottom (initial condition). This is because the potential energy is minimized (thus kinetic is maximized) when the large mass is at its low point. Therefore, the lest centripetal force will be when mass A is at the top (final condition). So now let's begin with the equation for conservation of energy:

Unfortunately we can't eliminate any terms, so let's begin expanding them one at a time:

If we assume that the lowest point of the circle has a height of 0, the initial potential energy of mass A will be 0:

We also know that the height at the top of the circle is exactly one rod length from our reference height, so we can say:

Moving on to initial kinetic:

Since bot velocities are always the same, we can say:

Now final kinetic. We'll just skip ahead to it's final form:

Note how it's the same as before, except that the masses are switched. No final kinetic. We'll skip to it's final form again:

Now plugging all of these back into the expression for conservation of energy, we get:

In order to find the final centripetal force, we are going to need to need the final velocity, so let's begin rearranging for that:

Factoring and dividing, we get:

We know every variable except initial velocity, which we can solve for by using the equation for centripetal force:

Where mass is mass B, v is the initial velocity, and radius is half the length of the rod:

Now we rearrange for initial velocity:

Plugging this back into our big equation along with the rest of the variable:

Now using the expression for centripetal force to find our final answer:

5

A car of mass is driving around a circular track of radius at a constant velocity of . The centripetal force acting on the car is . If the car's velocity is doubled, what is the new centripetal force required for the car to drive on the circular track?

Explanation

The equation for centripetal force is:

.

If is doubled and becomes , is quadrupled. Centripetal force is proportional to the square of velocity.

6

At the 2015 NCAA Track Championships, a competitor in the hammer is in the midst of his throw. During a period of his pre-throw spinning, his hammer is moving at a constant angular velocity of in its circular orbit. Being a regulation hammer, it weighs , and the length of its rope is . What is the magnitude of the force exerted towards the competitor along the hammer's rope?

Explanation

Because the hammer is moving at a constant angular velocity, the centripetal force along the hammer rope can be expressed as . By plugging in the numbers given by the problem we get which, when simplified gives us approximately or

7

The radius of earth is . Assume the Earth is perfectly spherical.

A space shuttle has a mass of approximately 2.04 million kilograms when ready for launch. What is the difference of the force needed to create liftoff if the shuttle were launched from the equator versus the North Pole (due to the rotation of the Earth)?

It would need less to liftoff.

It would need less to liftoff.

It would need the same upward force to liftoff.

It would need less to liftoff.

It would need less to liftoff.

Explanation

To find the difference in force upwards for liftoff, we need to find how much centrifugal force is generated from the rotation of the Earth. The equation for centrifugal force is:

Where is the mass of the object, is the distance from the center of rotation the object is, and , being the period of the rotation. We know that the earth rotates once per day, so we can find the period of rotation in seconds:

Now we solve for the centrifugal force:

This is all we need to calculate, because we know that the force pulling the shuttle down due to gravity would be identical from the equator to the North Pole on our idealized spherical Earth.

8

Consider the following system:

Spinning rod with masses at end

Two spherical masses, A and B, are attached to the end of a rigid rod with length l. The rod is attached to a fixed point, p, which is at a height, h, above the ground. The rod spins around the fixed point in a vertical circle that is traced in grey. is the angle at which the rod makes with the horizontal at any given time ( in the figure).

If the masses are traveling at a velocity of as they mass through the horizontal, what is the greatest centripetal force felt by mass B? Neglect air resistance and internal friction forces.

Explanation

Before we even begin thinking about what sort of equations we will be using, we first need to determine at which point mass B is feeling its greatest centripetal force. This will be whenever the mass is traveling at its fastest. Thinking about the law of conservation of energy, the mass will be traveling at its fastest when the potential energy of the system is minimized. There are two different masses in the system that are linked. Therefore, potential energy will be minimized when the larger mass is at its lowest point. This is when the rod is vertical and mass A is at the bottom. Now we can move on, beginning with the expression for centripetal force:

The only value we don't know is velocity. However, we can find it by using the expression for conservation of energy:

If we assume that point p is at a height of 0, we can eliminate initial potential energy to get equation (1):

Now let's expand each term moving from left to right:

Since the velocities of both masses are the same, we can say:

Moving onto final potential:

The final position of the rod is vertical with mass A at the bottom (which is a half length of rod below our reference point) and mass B at the top (half length rod above), so we can rewrite

Moving on to final kinetic:

Substituting these into equation (1) we get:

To make things look nicer, lets eliminate the fraction from each term:

We are solving for final velocity, so let's begin rearranging:

Note how the masses were switched in the final term since it switched sides. No some more rearranging:

We know all of those values, so time to plug and chug:

Now we go back to our expression for centripetal force:

We need to determine r, which is simply half the length of the rod:

Now that we know all of our values, we can solve the problem:

9

A person of mass is riding a ferris wheel of radius . The wheel is spinning at a constant angular velocity of . Determine the force exerted on the rider by their seat at the top of the ferris wheel.

None of these

Explanation

Convert to :

Definition of centripetal acceleration:

Plug in values:

Superposition of forces:

Where both and are pointing down, and thus negative.

Solve for

Plug in values:

10

Estimate the centripetal acceleration of a car driving around a long perfectly circular track.

Explanation

Convert to

Convert to :

Use the equation for centripetal acceleration:

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