All AP Physics 1 Resources
Example Questions
Example Question #41 : Centripetal Force And Acceleration
A model train completes a circle of radius in . Determine the centripetal acceleration.
None of these
One circle is equal to , thus the linear velocity is:
Using
Example Question #41 : Centripetal Force And Acceleration
A model train completes a circle of radius in . Determine the centripetal force.
None of these
One circle is equal to , thus the linear velocity is
Using
Use Newton's second law:
Example Question #81 : Circular And Rotational Motion
A model train completes a circle of radius in . Determine the angular momentum.
None of these
One circle is equal to , thus the linear velocity is:
The definition of angular momentum:
Plugging in values:
Example Question #131 : Circular, Rotational, And Harmonic Motion
An object has a mass M, velocity V, and moves in a circle with radius R.
What happens to the centripetal acceleration on the object when the mass is doubled?
The centripetal acceleration remains the same.
The centripetal acceleration remains the same.
The centripetal acceleration is halved.
The centripetal acceleration is doubled.
The centripetal acceleration is quadrupled.
The centripetal acceleration remains the same.
The equation for centripetal acceleration is:
Where is the centripetal acceleration on an object, is the velocity of an object, and is the radius in which the object moves in a circle.
We can see that mass does not play a role in the centripetal acceleration of an object, so no matter what happens to the mass, the centripetal acceleration remains the same.
Example Question #44 : Centripetal Force And Acceleration
An object has a mass M, velocity V, and moves in a circle with radius R.
What happens to the centripetal acceleration on the object when the radius is halved?
The centripetal acceleration of the object is quadrupled.
The centripetal acceleration of the object remains the same.
The centripetal acceleration of the object goes to zero.
The centripetal acceleration of the object is doubled.
The centripetal acceleration of the object is halved.
The centripetal acceleration of the object is doubled.
The equation for centripetal acceleration is:
Where is the centripetal acceleration on an object, is the velocity of an object, and is the radius in which the object moves in a circle.
The radius has an inverse relationship with centripetal acceleration, so when the radius is halved, the centripetal acceleration is doubled.
Example Question #42 : Centripetal Force And Acceleration
An object has a mass M, velocity V, and moves in a circle with radius R.
What happens to the centripetal acceleration on the object when the velocity is doubled?
The centripetal acceleration of the object is quadrupled.
The centripetal acceleration of the object remains the same.
The centripetal acceleration of the object is halved.
The centripetal acceleration of the object goes to zero.
The centripetal acceleration of the object is doubled.
The centripetal acceleration of the object is quadrupled.
The equation for centripetal acceleration is:
Where is the centripetal acceleration on an object, is the velocity of an object, and is the radius in which the object moves in a circle.
The velocity has an quadratic relationship with centripetal acceleration, so when the velocity is doubled, the centripetal acceleration is quadrupled.
Example Question #51 : Centripetal Force And Acceleration
It takes a disc player seconds to get up to its speed of rotations per second, what is the angular acceleration of the disc player in radians?
Similar to linear acceleration, angular acceleration is related to the angular velocity by the following
from here we simply plug in our values of and to get . Next to convert to radians we take the relations that pi is equal to rotation and multiply our by to arrive at our final answer of
Example Question #52 : Centripetal Force And Acceleration
You decide to ride an attraction at the fair; it is a circular, flat, ride that spins very fast; so fast, that it's nearly impossible to fall to the ground. Riders "stick" to the walls when the ride hits its top speeds.
Let's say the attraction spins in a clockwise manner. While riding the attraction, toward which direction is the centripetal (net) force pointing?
Outward
Inward
Against motion (counter clockwise)
With motion (clockwise)
Inward
Most of the success for this ride experience can be credited to Newton's Third Law. It states that if object 1 exerts onto object 2, then object 2 must be exerting a force . They are equal and opposite forces, called an action reaction pair. This concept applies when you're typing on your cell phone. when you touch down on your screen or button, your finger doesn't go through the phone, right? The phone is sending back an equal force to you which prevents your finger from going through the screen.
The same thing applies on the attraction. Your body is pushing outward on the walls of the ride. If no centripetal force was present, your body would project on a tangent as depicted in the diagram below. However, centripetal force is present. This is the force that is countering your body; it is pushing you towards the center. Now as we know, centripetal force isn't ever drawn on a diagram because it is a "net" force, or a collection of forces. But what we do know is that during circular motion, centripetal force (and acceleration) point towards the center.
During the ride, your body is trying to follow the tangental paths shown, but the reciprocal force, , is keeping your body stationary on the wall. The wall is following a circular path, which makes you follow the circular path. This action reaction pair is what allows the attraction to work. Centripetal force points inwards.
Example Question #1 : Torque
A bolt connecting the main and rear frame of a mountain bike requires a torque of to tighten. If you are capable of applying of force to a wrench in any given direction, what is the minimum length of the wrench that will result in the required torque?
The minimum length of the wrench will assume that the maximum force is applied at an angle of . Therefore, we can use the simplified expression for torque:
Here, is the length of the wrench.
Rearranging for length and plugging in our values, we get:
Example Question #1041 : Newtonian Mechanics
A uniform rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what distance from the left end of the rod should a 0.6kg mass be hung to balance the rod?
42cm
The rod cannot be balanced with this mass
45cm
48cm
50cm
45cm
The counterclockwise and clockwise torques about the pivot point must be equal for the rod to balance. Taking the fulcrum as the pivot point, the counterclockwise torque is due to the rod’s weight, gravitational force acting downwards at the center of the rod. If we use the pivot as our reference, then the center of the rod is 15cm from the reference.
Set this equal to the clockwise torque due to the additional mass, a distance r to the right of the pivot.
.
Solving for r gives r = 0.05m to the right of the pivot, so 40 + 5 cm from the left end of the rod.
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