Question

A disc of radius \(5~\text{cm}\) rolls on a horizontal surface with linear velocity \(v = 1 \hat{i}~\text{m/s}\) and angular velocity \(50 (-\hat{k})~\text{rad/sec}\) . Height of particle from ground on rim of disc which has velocity in vertical direction is (in \(\text{cm}\)) -

• A. 3.0
• B. 3.00
• C. 3

Answer 1

Answer: (A), (B), (C)

Let us represent the given situation pictorially, we get

Given:
$R=5\text{cm}=5\times {10}^{-2}\text{m},\omega =50\text{rad/sec}$
We have, $\theta$ such that :-
$v\cos \theta =v_{cm}=1$ and $v=R\omega (\because \text{Condition is of pure rolling})$
Thus, $R\omega \cos \theta =1$
$\Rightarrow 50\times 5\times {10}^{-2}\times \cos \theta =1$
$\Rightarrow (250\times {10}^{-2})\cos \theta =1$
$\Rightarrow \cos \theta =\frac{1}{2.5}=\frac{2}{5}$
Let the height of the particle be $h$
Therefore, $h=R-R\cos \theta$
$=R[1-\cos \theta ]$
$=5[1-\frac{2}{5}]$
$=3\text{cm}$