AP Physics 1 : Circular and Rotational Motion

Study concepts, example questions & explanations for AP Physics 1

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Example Questions

Example Question #2 : Centripetal Force And Acceleration

A ball of mass  is on a string of length . If the ball is being spun in vertical circles at a constant velocity and with a period of , what is the maxmium tension in the string?

 

Possible Answers:

Correct answer:

Explanation:

First, we need to identify at which point in the circle the string is experiencing the most tension. There are two total forces in the system: gravity and tension. It is important to note that the tension isn't only resulting from gravity; it also includes the centripetal force required to keep the ball in circular motion. Thinking practically, we can say that the greatest tension will be when the ball is at its lowest point (gravity and tension are in opposite directions). At this point we can write:

Expand our terms for force:

We know the acceleration due to gravity, but we need to determine the centripetal accerleration. The formula for that is:

We know the radius (length of the string), so we need to develop an expression for velocity. We can use the period and circumference of circle:

Here, we use  to denote period.

Substituting this into the expression for centripetal acceleration:

Substituting this back into the equation for tension, we get:

We have all of these values, allowing us to solve:

Example Question #1 : Centripetal Force And Acceleration

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?

Possible Answers:

Correct answer:

Explanation:

The equation for centripetal force is:

.

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

Example Question #5 : Centripetal Force And Acceleration

Consider a situation in which a 2kg block slides down a ramp and then around a circular loop with a radius of 12m. If friction between the surfaces is negligible, what is the minimum height that the block can start off at so that it will go all the way around the loop without falling off?

Possible Answers:

Correct answer:

Explanation:

To start off with, it is useful to consider the energy of the system. Initially, the block is at a certain unknown height, and will thus have gravitational potential energy.

Since we are assuming that there is no friction in this case, then we know that total mechanical energy will be conserved. Therefore, all of the gravitational potential energy contained in the block will become kinetic energy when it slides down to the bottom of the loop. But, once the block begins to slide up the loop, it will lose kinetic energy and will regain some gravitational potential energy. Therefore, at the top of the loop, the block will have a combination of kinetic and gravitational potential energy whose sum is equal to the initial energy of the system.

Due to conservation of energy, we can equate the two.

In addition to considering energy, it is also necessary to consider the forces acting on the block in this scenario. When at the top of the loop, the block will experience a downward force due to its weight, and another downward force due to the normal force of the loop on the block. Furthermore, because the block is traveling along a circular path while in the loop, it will experience a centripetal force. At the top of the loop, the centripetal force will be due to a combination of the weight of the block as well as the normal force.

We're looking for a starting height that will just allow the block to travel around the loop. The minimum height will be the height such that the block will just start to fall off. When falling off, the block will no longer be touching the loop and therefore, the normal force will be equal to zero. This is the situation we are looking for, and since the normal force is zero, only the block's weight will contribute to the centripetal force at the top of the loop.

The preceding expression gives us the value of velocity that will allow the block to have enough kinetic energy while at the top of the loop to not fall off. We can plug this expression into the previous energy equation.

Example Question #993 : Newtonian Mechanics

Tarzan problem

A person is crossing a canyon by swinging on a vine, as shown in the given figure. The person has a mass of 90kg and the vine he is using is 15m long from where it connects to the tree to the person's center of mass. At the instant he is at the lowest point of the swing, so the vine is going straight up from his hands, the person is moving at . At that instant, what is the tension force in the vine?

Possible Answers:

Correct answer:

Explanation:

Like all force problems, this one starts with a clear free body diagram:

 

Tarzan fbd

The tension  points along the vine (tensions can only pull), so it goes straight up. The force of gravity  points straight down, as it always does. The two do not add to zero, however, since the person is undergoing circular motion. Instead, they add to a net force pointing towards the center of the circle that the person is making, which is up at the place where the vine is attached to the tree. Draw a vector diagram:

Tarzan vd

Then write the equation about the lengths of the vectors: the length of the gravity vector plus the length of the net force vector equals the length of the tension vector:

The net force for an object undergoing circular motion is mass times speed squared divided by the radius of the circle.  is the gravity force constant. Some use , but the AP physics 1 test allows you to use , which makes it a lot easier. Plug in the numbers, and solve for the tension:

This answer is reasonable since the vine has to both hold the person up and provide a centripetal force; that is why the tension is is greater than his weight alone.

Example Question #3 : Centripetal Force And Acceleration

We can determine the centripetal force exerted by the nucleus on an electron. The diameter of an atom is , and an electron is moving at in a circular motion around the nucleus. The mass of an electron is roughly 

What is the centripetal force exerted by the nucleus on an electron? 

Possible Answers:

Correct answer:

Explanation:

The formula for centripetal force is:

Where  is the mass of the object,  is its velocity, and  is the radius of the circle made by the motion of the object around the center. We can model an atom like a circle, with its nucleus being its center. Since we are given the diameter of an atom, its radius will be

 

Example Question #1 : Centripetal Force And Acceleration

Imagine a car driving over a hill at a constant speed. Once the car has reached the apex of the hill, what is the direction of the acceleration?

Possible Answers:

No acceleration

In direction of motion

Downwards

Opposite direction of motion

Upward

Correct answer:

Downwards

Explanation:

If we imagine the hill as a semi-circle, it appears that the car is moving along a circle. At the apex of the hill, the car's acceleration points downwards as this points towards the center of the circle. If an object travels in a circular fashion, at a constant speed, the direction of acceleration is always towards the center of the circle. This type of acceleration arises do to the change in velocity. Although the speed is constant, the direction changes. 

Example Question #1041 : Ap Physics 1

The passengers in a car driving over a hill at a speed of   feel weightless.

 

Calculate the radius of curvature of the hill.

Possible Answers:

Correct answer:

Explanation:

We can solve this problem by looking at all of the forces acting on the car at the top of the circular path.

The weight is positive, because it points towards the center of the circle, while the normal force  is negative because it points away from the circle.

Since we are asked to look at the special case where the occupants of the car feel weightless, this tells us that the normal force experienced by the occupants must be zero.

 

Example Question #1041 : Ap Physics 1

Imagine a table that has a hole drilled through the center. Through this hole is a string that connects one mass that hangs under the table, to a mass that is on top of the table and spinning. mass M weight 10 kg, mass m weighs 5 kg and the radius, R, of the circle mass M traces out is 50 cm. At what speed, v, does the mass M need to rotate at in order to keep mass m suspended?

 

Possible Answers:

Correct answer:

Explanation:

For this problem, we recognize that the force keeping the mass m suspended is the tension in the string. This tension is balanced by the gravitational force on m such that m does not accelerate. So how do we relate the speed of M to the tension? We must ask ourselves why is there tension in the string at all? Any object that travels in a circle has some centripital acceleration acting on it that point towards the center of the circle (for constant speed). This force is equal to the mass time the centripial acceleration. 

I have stated previously that this force must be equal to the gravititational force acting on mass m. So,

Solving for v,

Plugging in values yields a speed of 1.6 m/s

Example Question #91 : Circular, Rotational, And Harmonic Motion

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

Possible Answers:

Correct answer:

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:

Example Question #1041 : Ap Physics 1

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?

Possible Answers:

Correct answer:

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 

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