A flywheel with moment of inertia 0.86 kg ·`m^2` and a cylinder of 5 cm radius of negligible mass are fixed to a common shaft (Fig.). A thread is wound around the cylinder, and a weight of 6.0 kg mass is attached to it. What time will the weight take to fall 1 m? What will be its final speed? Assume the initial speed to be zero.

Answer

`3.5s; 0.58 m//s`

Explanation

As the weight sinks, its potential energy transforms into the kinetic energies of the weight and of the rotating flywheel:
`mgh=1/2mv^2+1/2Iomega^2`
The flywheel and the cylinder rotate at the same angular velocity `omega = v//r` , therefore `2mgh = v^2 (m + 1//r^2)`. Hence
`v=sqrt((2mgh)/(m+I//r^2)`