A small mass M attached to a string slides in a circle (x) on a frictionless hor

Question

A small mass M attached to a string slides in a circle (x) on a frictionless horizontal table, with the force F providing the necessar tension (see figure). The force is then increased slowly and then maintained constant when M travels around in circle (y). The radius of circle (x) is twice the radius of circle Cy). XT M's kinetic energy at x is one quarter that at y. v M's angular velocity at x is one quarter that at y. As M moves from x to y, the work done by F is 0. M's angular momentum at x is that at y While going from x to y, there is a torque on M. Submit Answer Tries 0/20 Send Feedback Post Discussion

Explanation / Answer

As force is slowly increased and string is pulled down, Force acting on the string and displacement of the point where force is acting, are in same dircetion. Hence work done by force is +ve.

Force actng on the mass is the Tension in the string and acts along radial direction towards center. Hence torque of force on mass is zero.

As Torque on mass is zero there is no change in angular momentum of mass. So angular momentum at X is same as that at Y.

Angula Momentum L = Iw where I is moment of inertia and equal to MR2 and w is angular speed. Since angular momentum at X and Y are same

MRx2 wx = MRy2 wy

As Rx = 2Ry => wy = 4wx Or angular speed at X is quarter that at Y

Kinetic energy K = 1/2 ( I w2 ) = L2 /2I

Angular momentum L at X and Y is same, But 'I' at X is four times that at Y

Hence Ky = 4 Kx

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