After freezing a pendulum

A light metal rod of length L L is pivoted at one end. A hollow sphere filled with water has mass M M and radius R < L R<L is attached to the free end of the rod. The rod-sphere system performs angular SHM in vertical plane after being released from angle θ \theta with vertical as shown in the figure. Consider two cases

Case-1 : the water in the sphere is at 5 C 5^\circ C
Case-2 : the water in the sphere is at 2 C -2^\circ C

Which of the following statements are true? [Neglect the expansion of volume of water on freezing]

A) Restoring torque in case 1 = Restoring torque in case 2.
B) Restoring torque in case 1 < Restoring torque in case 2.
C) Angular frequency for case 1 > Angular frequency for case 2.
D) Angular frequency for case 1 < Angular frequency for case 2.
E) Angular frequency for case 1 = Angular frequency for case 2.

AE BC BE AC BD AD

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1 solution

Rajdeep Brahma
Jan 15, 2018

The torque is independent of moment of inertia which increases on freezing(since volume of ice is more than water it gets more spreaded and so more moment of inertia..the same reason why solid sphere has less moment of inertia than a hollow sphere)and hence w decreases .Torque is simply F×D .

Means friction is not there bw sphere and water layer. Thanks for question though I was not able to understand your solution firstly. A hollow sphere filled with water and rolling on inclined surface is also related to this problem....

Kunwar Pratap Singh - 3 years, 4 months ago

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actually I was trying to give logic behind the change in moment of inertia....why will it at all increase...then omega=root(c/I) u may use..be free if u have any doubt.

rajdeep brahma - 3 years, 4 months ago

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