All AP Physics 1 Resources
Example Questions
Example Question #1 : Circular Motion And Torque
A wheel of radius rolls along a flat floor and makes rotations over a period of time. What distance has the wheel traveled?
In order to find the distance the wheel travels, we need a way to convert angular displacement to linear displacement. We know that the circumference of a circle (or a wheel, in this case) is . This means that in one rotation, the wheel would travel a distance equal to its circumference.
Distance of one rotation equals: .
Since the wheel travels rotations, the total distance that the wheel travels will be equal to the distance traveled by one rotation multiplied by the number of rotations.
Distance of rotations equals:
Example Question #1 : Other Circular And Rotational Motion Concepts
A track runner who weighs 750 N is running around a curve at a speed of in an arc whose radius, , is 6.0 m. Find the centripetal force acting on him.
The equation for centripetal force on an object moving at a constant speed in a circular path is
The weight of the person is given to be . Assuming , we can find the mass of the person using the equation
We are given the velocity, , and the radius, . So, we can now solve for the centripetal force:
Example Question #2 : Other Circular And Rotational Motion Concepts
A 55 cm rope is tied to the handle of a bucket, which is then spun in a vertical circle. The mass of the bucket is 3.3 kg. What is the minimum speed required such that the rope does not become slack when the bucket reaches the highest point in the circle?
The faster that the object is moving while being spun, the greater the force of tension in the rope will be. Therefore, if we want to find the minimum speed of the bucket at the highest point in the circle such that the rope doesn't go slack, we should assume that the tension at that point is zero.
If the tension is zero, then the only force acting on the bucket at that point would be the force of gravity pulling down on the bucket. If the rope does not go slack, then the bucket must still be moving in a circular motion, so we can use the equation for circular/centripetal motion:
Since gravity is the only force, then
We can divide by on both sides to cancel that out, and then solve for :
The radius was given to us as 55 cm. We have to convert this to meters. Since , we should divide by 100 to get . Now, plug that into the equation above and solve for velocity:
Example Question #1 : Other Circular And Rotational Motion Concepts
Bob is riding a merry-go-round. He is 10 meters from the center and the merry-go-round completes two revolutions per minute. What is the centripetal acceleration acting on Bob?
The formula for centripetal acceleration is:
First, we need to solve for velocity by taking the distance Bob travels and dividing by the time it takes to travel that distance. The distance he travels is the circumference of the circle, and the time is 30 seconds, because the merry-go-round completes two revolutions per minute, or one revolution every thirty seconds:
Therefore,
Example Question #191 : Circular, Rotational, And Harmonic Motion
Without doing any calculations, what would reach the bottom of an incline first: a ring of mass , or a uniform density disk of mass ? It may assumed that they have the same radius and both start from rest.
The ring
Cannot be determined without knowing the angle of the incline
It will be a tie
The disk
Cannot be determined without knowing the density of the material
The disk
The ring has more of it's mass further from the center than the disk, which is more spread out. Moment of inertia increases if there is a larger "circle" that the mass has to make. Thus, if they have the same total mass, the disk will have a lower moment of inertia, thus it will reach the bottom faster. An example of this phenomenon is that of an ice skater holding in his/her arms and legs close to the center of rotation when doing jumping spins. If the ice skater were to spread his/her limbs out, their moment of inertia increases, and they would not be able to spin as fast.
Example Question #1 : Other Circular And Rotational Motion Concepts
In our solar system, it is known that the Earth moves around the Sun at a speed of about . In this case, is the earth accelerating?
Yes, because speed is not constant
No, because there is no acceleration
Yes, because velocity is not constant
No, because the speed is constant
Yes, because velocity is not constant
For this question, we have to consider the case of the earth orbiting the sun. Just like any scenario in which an object is moving in a circular motion, we know that there is a centripetal force that is constantly pulling the object inward. Thus, even though the object's speed may be constant, its velocity is constantly changing. This is because speed is a scalar that is not affected by direction. Velocity, on the other hand, is a vector which depends on direction.
Example Question #1 : Other Circular And Rotational Motion Concepts
Suppose that a mass hanging from a string attached to the ceiling is swinging in a circular motion, as shown in the diagram below.
If the mass is swinging at a constant speed of , and the mass is located from the center of its circular motion, what is the angle of the hanging mass from the ceiling?
There is not enough information given
To answer this question, we'll need to first identify all the forces acting on the hanging mass. To do this, we have to consider the forces acting along the y-axis and x-axis.
First, let's consider the forces in the y-direction. Since the mass is not moving in the y-direction, we know that the net force acting on the mass in the vertical direction is zero. We know that the weight of the mass acts downward, and the vertical component of the string's tension acts upward.
Next, let's look at the forces acting on the mass in the horizontal direction. Since the mass is moving in a uniform circular motion, we know that there is a centripetal force associated with this. Moreover, we know that the x-direction of the tension in the string is acting in the x-direction. Thus, this component of the string's tension must be providing the centripetal force.
Now, if we divide these two equations, we can cancel out the terms for tension and mass.
Further rearrangement of the above expression allows us to solve for the angle.
Example Question #4 : Other Circular And Rotational Motion Concepts
Suppose that an astronaut is traveling around the Earth at a constant speed. Which of the following statements regarding the astronaut's circular motion is correct?
None of these
The centripetal acceleration provided by the Earth's gravitational field is constant
If the astronaut's speed is doubled, his centripetal acceleration will also double
If the distance between the astronaut and the Earth were to double, the centripetal acceleration of the astronaut would also double
The astronaut is constantly in a state of free-fall
The astronaut is constantly in a state of free-fall
For this question, we're asked to evaluate a true statement regarding circular motion. Hence, we'll need to look at each answer choice individually and consider whether it is true.
- The centripetal acceleration provided by the Earth's gravitational field is constant
This is not a true statement. Remember that acceleration is a vector. Even though the astronaut's centripetal acceleration is constant in magnitude, it is constantly changing direction as the force will always point towards the center of the Earth.
- If the astronaut's speed is doubled, his centripetal acceleration will also double
This statement is false. Remember the expression for centripetal force.
Hence, if the astronaut's speed becomes twice as great, then his acceleration will become four times as great.
- If the distance between the astronaut and the Earth were to double, the centripetal acceleration of the astronaut would also double
This statement is not true. Again, we need only look to the expression shown above for centripetal force to see why. Because the centripetal acceleration and distance are both inversely proportional to one another, doubling the distance of the astronaut from the Earth would cause the acceleration to become half as great. Thus, it would decrease, not increase.
- The astronaut is constantly in a state of free-fall
This is a true statement, and thus it is the correct answer. Objects that are orbiting the Earth experience a centripetal force due to gravity. If we picture a snapshot of the movement of the astronaut, he will be moving tangentially to his circular displacement. As the astronaut moves tangentially, the force of gravity serves as the centripetal force that "pulls" him back, such that the astronaut does not move down the tangential path. This continuous "pull" away from the tangential path at each instant essentially means that our astronaut will constantly fall towards the earth while maintaining his circular path, and is thus constantly in a state of free-fall, where he will be weightless.
Example Question #2 : Other Circular And Rotational Motion Concepts
Two cars are racing side by side on a circular race track. Which has the greater linear velocity?
The car on the outside
The car on the inside
Impossible to determine
They have the same linear velocity
The car on the outside
If the cars are racing side by side on a circular track, then they have the same angular velocity, because they complete their circles in the same amount of time. However, the car on the outside has a larger circle to complete, and thus must have a higher linear velocity.
Example Question #5 : Other Circular And Rotational Motion Concepts
Use energy conservation and your knowledge of rotational dynamics to solve the following problem
For a ball of radius and mass starting at the top of a friction-less ramp of height , as shown below, what will the final velocity of the ball be just as it leaves the bottom of the ramp? Assume that the gravity on earth is and round your answer to the nearest 2 significant figures.
For this problem the ball picks up rotational velocity as well as linear velocity. Keeping this in mind we first set up our equation of conservation of energy
Because the object starts at rest the initial kinetic energy is 0 and likewise the final potential energy is 0 because it has spent all its potential energy to become kinetic.
Where m is mass of the ball, g is gravity, h is height of the ramp, v is final velocity of the ball, I is the moment of inertia of the ball, and w is the final rotational velocity of the ball.
Now because the ball is a solid sphere we can represent I and w as follows
and
Plugging this into our equation of energy conservation
Now solve for velocity:
Now plugging in our value for height, we arrive at a final answer of