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

Step-by-step 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* 9.8* 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)`