A mass m supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?

Question

A mass m supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release? (A) 2g/3 (B) g/2 (C) 5g/6 (D) g

Tardigrade Admin · Accepted Answer

For the mass m, mg-T=ma As we know, a=R α So, mg-T=mRα .....(i) Torque about centre of pully T x R = mR2α .....(ii) From Eqs. (i) and (ii), we get, a=g/2 Hence, the acceleration with the mass of a body fall is g/2.

Q. A mass m supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?

2g/3

g/2

5g/6

g

For the mass m, $m g - T = ma$
As we know, $a = R α$
So, $m g - T = m R α$ .....(i)
Torque about centre of pully
$T x R = m R^{2} α$ .....(ii)
From Eqs. (i) and (ii), we get, $a = g /2$
Hence, the acceleration with the mass of a body fall is g/2.

Q. A mass m supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?

2g/3

g/2

5g/6

g

For the mass m, mg−T=ma As we know, a=Rα So, mg−T=mRα .....(i) Torque about centre of pully TxR=mR2α .....(ii) From Eqs. (i) and (ii), we get, a=g/2 Hence, the acceleration with the mass of a body fall is g/2.

For the mass m, $m g - T = ma$
As we know, $a = R α$
So, $m g - T = m R α$ .....(i)
Torque about centre of pully
$T x R = m R^{2} α$ .....(ii)
From Eqs. (i) and (ii), we get, $a = g /2$
Hence, the acceleration with the mass of a body fall is g/2.