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). If you look up this question and look at the other pics it is the same image.

Pick one for each : True, false, greater than, less than, equal to.

1.)M's kinetic energy at x is half that at y.

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

3.) M's angular velocity at y is four times that at x.

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

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

Explanation / Answer

F = Mv² / R is the centripetal force at circle Y

F' = Mv² / 2R is the centripetal force at circle X

angular momentum is conserved since there is no external force or torque

=> I = I -------------------------------------> (i)

I = M(2R)² = 4MR²

I = M(R)² = MR²

or I = 4I -------------------------------> (ii)

from (i) and (ii) => = 4 -------------------------------(iii)

=> v = R = R.4 = 2v ------------------------------------ (iv)

comparing kinetic energies:

k.e / k.e = ½mv² / ½mv² = (2v)² / v² = 4/1 -------------> from (iv)

=> k.e = 4k.e -------------------->(v)

therefore Answers are:

M's angular velocity at Y is four times that at X.
True.

While going from Y to X, there is a torque on M
False: as the only force acting on M is along the radius => = Fx0 = 0

M's kinetic energy at Y is twice that at X.
False. its 4 times.

As M moves from Y to X, the work done by F is .... 0.
< 0 as it decreases the k.e of M from 4k.e to k.e

M's angular momentum at Y is .... that at X.
Equal.

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