## Saturday, October 06, 2007

### Rotational Motion of a Reel – Two Multiple Choice Questions

The following question on the acceleration of a reel carrying stitching thread is an interesting one:

A reel of mass ‘m’ in the form of a short solid cylinder carries light inextensible string wound round it. The free end of the string is tied to a hook fixed to the ceiling and the reel is allowed to roll down, unwinding the string. The linear acceleration of the reel is

(a) g (b) g/3 (c) g/2 (d) 3g/2 (e) 2g/3

If ‘a’ is the acceleration of the of the reel we have

Mg – T = ma

where ‘M’ is the mass of the reel and ‘T’ is the tension in the string.

Therefore, T = mg – a

[You can directly write the above value of T if you remember that T is equal in magnitude to the weight of the reel while moving down with acceleration ‘a’ as in a lift]

If the radius of the reel is ‘R’, the torque on the reel is TR.

Therefore, we have

TR = Iα

where ‘I’ is the moment of inertia of the reel ( I = MR2/2) and ‘α’ is its angular acceleration, which is equal to a/R.

The above equation therefore becomes

(Mg – a)R = (MR2/2) (a/R)

This gives a = 2g/3.

A solid cylinder having mass M and radius R is free to rotate about its axis and has a light inextensible string wound roun it. A body of mass M/2 is attached to the free end of the string. If the system is released from rest, the mass will move down with a linear acceleration of

(a) g (b) g/3 (c) g/2 (d) 3g/2 (e) 2g/3

The tension in the string is given by

T = (M/2)(g – a).

Since TR = Iα, we have

(M/2)(g – a)R = (MR2/2) (a/R).

This equation yields a = g/2.

[It will be interesting to note that if the mass suspended were M, the acceleration would have been 2g/3, which is the acceleration of the falling reel in the previous question].