Question

A 7.50 kg disk with a diameter of 88.0 cm is spinning at a rate of 280 revolutions per minute. A block of wood is pressed into the edge of the disk with a force of 35.0 N. The coefficient of friction between the wood and the disk is 0.300. How long does it take the disk to stop spinning?

Answer 1

Time taken by disk to stop spinning will be 4.60 sec

Step-by-step explanation:

It is given mass m = 7.50 kg

Diameter of the disk d = 88 cm = 0.88 m

So radius of the disk
`r=(d)/(2)=(0.88)/(2)=0.44m`

Angular speed of the disk
`\omega =(2\pi N)/(60)=(2* 3.14* 280)/(60)=29.30rad/sec`

Force is given F = 35 N

Coefficient of friction
`\mu =0.3`

Friction force will be equal to
`f=\mu F=0.3* 35=10.5N`

So torque will be equal to
`\tau =Fr=10.5* 0.44=4.62Nm`

Torque is also equal to
`\tau =I\alpha`

Moment of inertia of the disk
`I=(1)/(2)Mr^2=(1)/(2)* 7.5* 0.44^2=0.726kgm^2`

So
`4.62=0.726* \alpha`

`\alpha =6.964rad/sec^2`

So time taken will be equal to
`t=(\omega -0)/(\alpha )=(29.30-0)/(6.964)=4.60sec`