Question

A disc of radius $0.2\text{m}$ is rolling with slipping on a flat horizontal surface as shown in the figure. The instantaneous centre of rotation is (the lowest contact point is O and centre of disc is C)

• A. zero
• B. $0.1\text{m}$ above O on line OC
• C. $0.2\text{m}$ below O on line OC
• D. $0.2\text{m}$ above O on line OC

Answer 1

The correct option is C $0.2\text{m}$ below O on line OC

Given, $v=4m/s$ and tangential velocity = $\omega x$
let take two point $P$ and $Q$ at vertically upward and vertically downward.

In a point P both velocity $\left(\omega x\right)$ and $v$ are in a same direction hence system will never be in rest. So, this is wrong.

Now, for point $Q$

$\omega x=4$ (given, $\omega =10red/sec$)
$x=\frac{2}{5}=0.4\neg m$
and given radius of sphere =$0.2m$
the distance from the point of contact= $0.4-0.2=0.2m$

Hence, The instantaneous centre of rotation is $0.2\text{m}$ below O on line OC.