Question

A wheel having radius `10cm` is coupled by a belt to another wheel of radius `30cm`. 1st wheel increases its angular speed from rest at a uniform rate of `1.57 rad//s^2`. The time for 2nd wheel to reach a rotational speed of `100 rev//min` is …(assume that the belt does not slip)

Answer 1

As the belt does not slip `v_(A)=v_(C)`
i.e., `r_(A)omegaA=r_(C)omega_(C)` [as `v=romega`]
Here, `r_(A)=10cm, r_(C^(-))=30cm`
and `omega_(C)=(2pixx100//60)` rad/s
So, `10omega_(A)=30xx(2pixx100)/(60)`
i.e., `omega_(A)=10pi` rad/s
Now, from 1st equation of rotational motion, i.e.,
`omega=omega_(0)+alphat`
`t=(omega)/(alpha)`
i.e., `t=(10xxpi)/(1.57)=20` sec.