MCAT Physical : Rotational, Circular, and Harmonic Motion
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Example Questions
Example Question #1 : Centripetal Force And Acceleration
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, 
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 
An object moving linearly at constant velocity has 0 acceleration, so 
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.
Example Question #1 : Centripetal Force And Acceleration
What is the approximate centripetal acceleration of a point on the rim of a bicycle wheel with a diameter of 
Centripetal acceleration is given by the equation:
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.
Now we can convert the angular velocity to the linear velocity using the radius.
Use the radius and the linear velocity to calculate the centripetal acceleration from the original equation.
Example Question #3 : Centripetal Force And Acceleration
What is the centripetal force on a 
Centripetal force is given by the equation:
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.
Example Question #4 : Centripetal Force And Acceleration
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?
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 
Example Question #1 : Centripetal Force And Acceleration
A man of mass 
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:
Rearranging for final velocity:
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:
Now we can plug in our values and solve:
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:
Now that we know the force in Newtons, we can convert this value to g's.
For this man:
Therefore,
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