# A solid cylinder rolls down an inclined plane of height 3 m and reaches the bottom of plane with angular velocity of 2√2 rad/s. The radius of the cylinder must be ?

## [Take g=10 m/s^2] a. 5 cm , b. 0.5 m, c.√10 cm , d.√5 m(answer), e.10 cm How do you get this answer? Thanks! Dec 21, 2016

When the solid cylinder rolls down the inclined plane, without slipping, its total kinetic energy is given by
$K E \text{due to translation"+"Rotational " KE=} \frac{1}{2} m {v}^{2} + \frac{1}{2} I {\omega}^{2}$ ....(1) If $r$ is the radius of cylinder,
Moment of Inertia around the central axis $I = \frac{1}{2} m {r}^{2}$ .....(2)
Also given is $\omega = \frac{v}{r}$ ....(3)

Assuming that it starts from rest and ignoring frictional losses, at the bottom of the plane
Total kinetic energy is $= \text{Potential Energy at the top of plane} = m g h$ .....(4)
Using Law of conservation of energy and equating (1) and (4) we get

$m g h = \frac{1}{2} m {v}^{2} + \frac{1}{2} I {\omega}^{2}$ ......(5)

Using (3) and (4), equation (5) becomes
$m g h = \frac{1}{2} m {\left(r \omega\right)}^{2} + \frac{1}{2} \times \frac{1}{2} m {r}^{2} {\omega}^{2}$
$\implies g h = \left(\frac{1}{2} + \frac{1}{4}\right) {r}^{2} {\omega}^{2}$
$\implies g h = \frac{3}{4} {r}^{2} {\omega}^{2}$
Solving for $r$
$\implies r = \sqrt{\frac{4 g h}{3 {\omega}^{2}}}$
Inserting given values we get value of radius $r$ as

r=sqrt((cancel4xx10xxcancel3)/(cancel3(cancel2sqrt2)^2)
$r = \sqrt{5} m$