An air puck of mass 0.664 kg is tied to a string and allowed to revolve in a cir

Question

An air puck of mass 0.664 kg is tied to a string and allowed to revolve in a circle of radius 1.26 m on a frictionless horizontal table. The other end of the string passes through a hole in the center of the table, and a mass of 1.10 kg is tied to it as seen in the figure. The suspended mass remains in equilibrium while the puck on the tabletop revolves.

A) What is the tension in the string?

B)What is the magnitude of the force which causes the centripetal acceleration of the puck?

C) What is the speed of the puck?

Explanation / Answer

a). Since the mass of 1.1 kg is in equilibrium.

Thus tension in string= Force on the mass due to gravity

Tension in the string, T= mg= (1.1)(9.8)= 10.78 N

b). Since whole system is in equilibrium, Thus the centripetal force due to revolving of air puck will be balanced by the tension in the string.

Thus Force due to centripetal acceleration = Tension in the string= 10.78 N.................(1)

c). Also we know that force due to centripetal acceleration, F= m'v2/r

  or by using equation 1,

   m'v2/r= 10.78

   v2 = (10.78 * 1.26)/0.664= 20.456

or v= 4.522 m/s

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