Question

A disk-shaped grindstone of mass 3.0 kg and radius 8.0 cm is spinning at 600 rev/min. After the power is shut off, a man continues to sharpen his axe by holding it against the grindstone until it stops 10 s later What is the average torque exertes by the axe on the grindstone?

Answer 1

τ=0.060 N.m

Step-by-step explanation:

By kinematics:

`\omega f = \omega o-\alpha*t`

Solving for α:

`\alpha=(\omega o-\omega f)/(t)`

where ωo = 600*2*π/60; ωf = 0; t=10s

`\alpha=6.283rad/s^2`

The sum of torque is:

`\tau=I*\alpha`

`\tau=M*R^2/2*\alpha`

`\tau=0.060 N.m`