Question

A uniform disk with a mass of 5.0 kg and diameter 30 cm rotates on a frictionless fixed axis through its center and perpendicular to the disk faces. A uniform force of 4.0 N is applied tangentially to the rim of the disk. What is the angular acceleration of the disk?

Answer 1

Answer:
`\alpha =10.66 rad/s^2`

Step-by-step explanation:

Given

mass of disk
`m=5 kg`

diameter of disc
`d=30 cm`

Force applied
`F=4 N`

Now this force will Produce a torque of magnitude

`T=F\cdot r`

`T=4\dot 0.15`

`T=0.6 N-m`

And Torque is given Product of moment of inertia and angular acceleration
`(\alpha )`

`T=I\cdot \alpha`

Moment of inertia for Disc
`I= (Mr^2)/(2)`

`I=0.05625 kg-m^2`

`0.6=0.05625\cdot \alpha`

`\alpha =10.66 rad/s^2`