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Example Questions
Example Question #11 : Circular Motion
A penny is placed 12.0cm from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly increased, the penny remains fixed on the turntable until a rate of 40 revolutions per minute is reached, at which point the penny slides off. Calculate the coefficient of static friction between the penny and the turntable.
Known
which should be converted to meters ()
which should be converted to
We are dealing with circular motion so the first thing to do is to identify the force that is causing the centripetal motion. In this case it is the force of friction which is holding the penny in place. Therefore we know that the centripetal force is equal to the force of friction.
The centripetal force is equal to the mass times the centripetal acceleration.
To calculate the centripetal acceleration you will need to know the velocity of the penny and the radius that the penny sits at.
Therefore
We also know that the frictional force is related to the coefficient of friction and the normal force.
In this case the penny is sitting on the surface and the gravitational force is equal in size to the normal force. Therefore
The force of gravity is equal to the mass times the gravitational acceleration.
We can put all this together to a larger equation that says that
Notice that mass falls out of both sides of the equation
We can now rearrange and solve for the coefficient of friction
Now we can plug in our known values
Example Question #11 : Understanding Circular Motion
A ball attached to a string is moving counterclockwise in a vertical circle. If the string is cut exactly at the point where the ball is at the top of its motion (the top of the circle), what direction will the ball move in initially?
Upward
Downward
Left
Right
Left
In circular motion, velocity is tangential to the circular path. Since the object is moving counterclockwise, at the top of the circle this tangent line points to the left. It may help to draw a diagram to better visualize this motion.
Example Question #13 : Circular Motion
Conceptual
Two children sit on a merry-go-round. One sits from the center, and the other sits from the center. If the children are in a straight line form the center, which child has a greater speed?
The child from the center
We must know the acceleration to draw a conclusion
We must know the period to draw a conclusion
The child from the center
They have the same speed
The child from the center
Velocity is given by distance per unit time.
When moving in a circle, the distance is the circumference, and each rotation takes exactly one period. We can substitute into the velocity formula.
If the children are in a straight line, that means that their periods (how long it takes to make one revolution) will be the same. The only thing that changes is r, the distance from the center. Since radius is in the numerator, we can conclude that increasing the distance from the center will increase the velocity.
Example Question #12 : Understanding Circular Motion
Two children sit on a merry-go-round. One sits from the center, and the other sits from the center. If the children are in a straight line from the center, which child has a greater speed?
The child from the center
We must know the period to draw a conclusion
We must know the acceleration to draw a conclusion
They have the same speed
The child from the center
The child from the center
Velocity is given by distance per unit time.
When moving in a circle, the distance is the circumference, and each rotation takes exactly one period. We can substitute into the velocity formula.
If the children are in a straight line, that means that their periods (how long it takes to make one revolution) will be the same. The only thing that changes is , the distance from the center. Since radius is in the numerator, we can conclude that increasing the distance from the center will increase the velocity.
Example Question #11 : Circular Motion
A rider on a Ferris wheel moves in a vertical circle of radius r at constant speed v. Is the normal force that the seat exerts on the rider at the top of the wheel?
The same as the force the seat exerts at the bottom of the wheel
Less than the force the seat exerts at the bottom of the wheel
More than the force the seat exerts at the bottom of the wheel
Less than the force the seat exerts at the bottom of the wheel
The centripetal force is what is acting on the rider. At the top of the Ferris wheel, the normal force is pointing up, and the gravitational force is pointing down. The sum of these two forces must equal the centripetal force pointing downward toward the center of the circle. Therefore the normal force must be smaller than the gravitational force. At the bottom of the Ferris wheel, the same forces are present. However, the sum of these forces must equal the centripetal force point upward toward the center of the circle. Therefore the normal force must be greater than the gravitational force. Since the normal force must be greater than the gravitational force at the bottom and less than the gravitational force at the top, the force at the bottom must be greater than the force on the top.
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