Question

A pulley 12 cm in diameter is free to rotate about a horizontal axle. A 220-g mass and a 470-g mass are tied to either end of a massless string, and the string is hung over the pulley. Assuming the string doesn’t slip, what torque must be applied to keep the pulley?

Answer 1

Answer:0.147 N-m

Explanation:

Given

Diameter of Pulley $d=12 cm$

radius $r=6 cm$

mass of first object $m_1=220 gm$

mass of second object $m_2=470 gm$

Now both masses will exert a torque a on Pulley

Torque due to first Pulley $T_1=m_1g\cdot r$

$T_1=0.22\times 9.8\times 0.06=0.129 N-m$

Torque due to second mass on Pulley $T_2=m_2g\cdot r$

$T_2=0.276 N-m$

Total Torque by masses $T_{net}=T_2-T_1$

$T_{net}=0.276-0.129=0.147 N-m$

so we need to apply a torque of magnitude 0.147 N-m opposite to the direction of $T_{net}$