Question

A wooden disk of mass m and radius r has a string of negligible mass is wrapped around it. If the disk is allowed to fall and the string is held so that it unravels without sliding as the disk is falling, what will be the wooden disk's acceleration? What would be the tension on string in the process?

Answer 1

`a = (2)/(3)g`

`T = (mg)/(3)`

Step-by-step explanation:

As the disc is unrolling from the thread then at any moment of the time

We have force equation as

`mg - T = ma`

also by torque equation we can say

`TR = I\alpha`

`TR = (1)/(2)mR^2((a)/(R))`

`T = (1)/(2)ma`

Now we have

`mg - (1)/(2)ma = ma`

`mg = (3)/(2)ma`

`a = (2)/(3)g`

Also from above equation the tension force in the string is

`T = (1)/(2)ma`

`T = (mg)/(3)`