Question 28: Work, Energy and Power & Circular Motion

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 $g = 10\ \mathrm{m/s^2}$)

Accepted Answer

For the mass to remain stuck to the inner wall without falling, the upward
frictional force
must balance its weight.
Step 1: Forces acting on the mass
Weight downward:
$$mg$$
Normal reaction by the wall toward the centre:
$$N$$
Friction upward:
$$f$$
At the
minimum rotational speed
, the friction is just enough to prevent slipping, so
$$f_{\max} = \mu N = mg$$
Step 2: Normal reaction provides centripetal force
Since the drum is rotating, the normal force provides the centripetal force:
$$N = m\omega^2 r$$
Step 3: Use the condition for minimum speed
So,
$$\mu N = mg$$
Substitute $N = m\omega^2 r$:
$$\mu (m\omega^2 r) = mg$$
Cancel $m$:
$$\mu \omega^2 r = g$$
Hence,
$$\mu = \frac{g}{\omega^2 r}$$
Step 4: Substitute values
Given:
$g = 10\ \mathrm{m/s^2}$
$\omega = 5\ \mathrm{rad/s}$
$r = 4\ \mathrm{m}$
$$\mu = \frac{10}{(5)^2 \cdot 4} = \frac{10}{25 \cdot 4} = \frac{10}{100} = 0.1$$
Answer:
$$\boxed{0.1}$$
So, the correct option is
A
.

Answer

0.1

Answer

0.5

Answer

0.7

Answer

0.3

Work, Energy and Power & Circular Motion

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