An air puck of mass 0.612 kg is tied to a string and allowed to revolve in a cir

Question

An air puck of mass 0.612 kg is tied to a string and allowed to revolve in a circle of radius 1.27 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 1.36 kg is tied to it as seen in the figure.

The suspended mass remains in equilibrium while the puck on the tabletop revolves. What is the tension in the string?
What is the magnitude of the force which causes the centripetal acceleration of the puck?
What is the speed of the puck?

Explanation / Answer

the suspendeed mass M = 1.36 kg is in equilibrium

when a body is in equilibrium the net force acting = 0

net force acting on M = Fnet = T - M*g

T - tension in upward direction

M*g = weight in down wward direction

but Fnet = 0

T = M*g = 1.36*9.81 = 13.34 N   <<_---------answer

the tension in the string provides the necessary centripetal force Fc

Fc = T = 13.34 N <-answer

the puck is rotating at aradius r = 1.27 m

centripetal force Fc = m*v^2/r

v = sqrt(Fc*r/m)

v = sqrt((13.34*1.27)/0.612)

speed = v = 5.26 m/s <<<<<-------------answer

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