Laws of Motion - Result Question 53
53. A block of mass $10 kg$ is in contact against the inner wall of a hollow cylindrical drum of radius $1 m$. The coefficient of friction between the block and the inner wall of the cylinder is 0.1 . The minimum angular velocity needed for the cylinder to keep the block stationary when the cylinder is vertical and rotating about its axis, will be : $(g=10 m / s^{2})$
[2019]
(a) $\sqrt{10} rad / s$
(b) $\frac{10}{2 \pi} rad / s$
(c) $10 rad / s$
(d) $10 \pi rad / s$
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Answer:
Correct Answer: 53. (c)
Solution:
- (c) Given mass of block, $m=10 kg$; radius of cylindrical drum, $r=1 m$; coefficient of friction between the block and the inner wall of the cylinder $\mu=0.1$;
Minimum angular velocity $\omega _{\min }$
For equilibrium of the block limiting friction
$f_L \geq mg$
$\Rightarrow \mu N \geq mg$
$\Rightarrow \mu r \omega^{2} \geq mg$
Hrere, $N=mr \omega^{2}$
or, $m \geq \sqrt{\frac{g}{r \mu}}$
or, $\omega _{\min }=\sqrt{\frac{g}{r \mu}}$
$\therefore \omega _{\min }=\sqrt{\frac{10}{0.1 \times 1}}=10 rad / s$
