Question

You have a grindstone (a disk) that is 90.0 kg, has a 0.340-m radius, and is turning at 90.0 rpm, and you press a steel axe against it with a radial force of 20.0 N. (a) Assuming the kinetic coefficient of friction between steel and stone is 0.20, calculate the angular acceleration of the grindstone. (

Answer 1

Step-by-step explanation:

Given mass of grindstone
`m=90\ kg`

radius of stone
`r=0.34\ m`

angular speed of disc
`\omega =90\ rpm`

Steel Axle applying a force of
`F=20\ N`

coefficient of kinetic friction
`\mu =0.2`

Frictional Torque applied by steel is given by

`\tau=r* f_r`

`\tau =r* \mu F`

where
`f_r`=frictional force

`\tau =r* \mu * F`

`\tau =0.34* 0.2* 20`

`\tau =1.36\ N-m`

Torque is also given by

`\tau =I\cdot \alpha`

where
`\alpha`=angular acceleration

I=moment of Inertia

`\tau =0.5Mr^2* \alpha`

`0.5Mr^2* \alpha =1.36\ N-m`

`0.5* 90* 0.34^2* \alpha =1.36`

`\alpha =0.261\ rad/s^2`