A 0.5 kg mass is in contact against the inner wall of a cylindrical drum of radius 4 m rotating about its vertical axis

Question

A 0.5 kg mass is in contact against the inner wall of a cylindrical drum of radius 4 m rotating about its vertical axis. The minimum rotational speed of the drum to enable the mass to remain stuck to the wall (without falling) is 5 rad/s. The coefficient of friction between the drum’s inner wall surface and mass is ______. (Take g = 10 $m / s^{2}$ ) [2026]

Smart Achievers · Accepted Answer

(1)

$F.B.D of the block:$

$Considering F.B.D,$

$N = m ω^{2} R and f_{s} = m g$

$f_{s} \leq f_{l} \Rightarrow m g \leq μ N$

$m g \leq μ m ω^{2} R \Rightarrow μ \geq \frac{g}{ω^{2} R}$

$μ_{\min} = \frac{g}{ω^{2} R} = \frac{10}{25 \times 4} = \frac{10}{100} = 0.1$

Solution

1 0.1

2 0.5

3 0.7

4 0.3

Ans.
(1)

$F.B.D of the block:$

$Considering F.B.D,$

$N = m ω^{2} R and f_{s} = m g$

$f_{s} \leq f_{l} \Rightarrow m g \leq μ N$

$m g \leq μ m ω^{2} R \Rightarrow μ \geq \frac{g}{ω^{2} R}$

$μ_{\min} = \frac{g}{ω^{2} R} = \frac{10}{25 \times 4} = \frac{10}{100} = 0.1$

Want Us to Email you About Special Offers & Updates?