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 necessary 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 (y).

A)    M's kinetic energy at x is one quarter that at y.

B)    M's angular momentum at y is .... that at x.

C)    M's angular velocity at x is half that at y.

D)    While going from x to y, there is no torque on M.

E)    As M moves from x to y, the work done by F is .... 0.

Explanation / Answer

Ans:

1)False

2)Equal

3)False

4)False (No torque acting)

5)Greater than( Energy Increases)

Explanation

Relation between Angular momentum L and Moment of inertia I

Angular momentum  L = I * w,
I is the moment of inertia, which, for a point mass M rotating in circular motion I = M * r^2
In above problem
Ix = M * x^2 and Iy = M * y^2
and x = 2y; therefore
Iy = M * y^2 = M * (x / 2)^2 = (1/4) * M * x^2
This gives relation Ix =4Iy

For angular momentum we have relations:

Lx = Ix * Wx and, from
Ly = Iy * Wy = [(1/4) * Ix] * Wy
Lx = Ly (conservation of angular momentum)

Hence
Wy = 4 * Wx

The rotational kinetic energy KE is given by:
KE = (1/2) * I * w^2

In this problem
KEx = (1/2) * Ix * wx^2 and
KEy = (1/2) * Iy * wy^2

Using previous relations we obtain that
KEy = (1/2) * ((1/4) * Ix) * (4 * wx)^2

= 4^2 / 4 * KEx = 4*KEx

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