Question
Question: A coin is placed on a rotating table. It is observed that the coin just slips when placed at a dista...
A coin is placed on a rotating table. It is observed that the coin just slips when placed at a distance of 4r from the center. On doubling the angular velocity of the table, it will just slip when its distance from the center equal to:
A: 2r
B: 4r
C: r
D: 4r
Solution
The condition when the coin just slips is the condition when the centrifugal force has just overcome the static friction or we can say just friction and coin is just about to slip in this case the value of centripetal force will be equal to the frictional force.
Complete step by step answer:
Here in this question, it is said that a coin is placed at a distance 4r from the center on the rotating table the angular velocity of the table is ω and the coin just slips this means at this instance the centrifugal force will be equal to frictional force.
As we know the centrifugal force is given by F=mω2r
The maximum frictional force is given by Fk=μkmg
∵F=Fk
∴Fk=mω2r
Centrifugal force, in this case, will be as r=4r so, Fk=4mω2r (1)
Now it is said that the angular velocity of the rotating table is increased to twice the initial value or
The new angular velocity of the table is 2ω
Now it is asked at what distance should the coin be placed so it just slips
Let the position be at x distance from the center
Now the centrifugal force at this point will be given by
F=mω2r
∵ω=2ω here so
F=4mω2x
Now when the coin just slips then this centrifugal force will be equal to the friction force
∵F=Fk
∴Fk=4mω2x (2)
Equating both (1) and (2) we get
4mω2r=4mω2x
→r=x
Final answer: (C) the coin should be placed at a distance r from the center is the angular velocity of the table is doubled.
Note:
-The maximum friction force will always be constant.
-If the coins slip then the centrifugal force should be equal to the maximum friction force.
-Centrifugal force always acts in a tangentially outward direction.