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 \( g = 10 \, \text{m/s}^2 \))}
Step 1: Understanding the Concept:
For a mass to remain stuck to the inner wall of a rotating cylinder, the upward frictional force must balance the downward gravitational force.
The frictional force is provided by the normal force, which in turn is generated by the centripetal acceleration of the rotating drum. Step 2: Key Formula or Approach:
Normal force exerted by the wall on the mass: \(N = m r \omega^2\).
Static friction force: \(f_s \le \mu_s N\).
For the mass to not fall, the friction must counteract gravity: \(f_s \ge mg\).
At the minimum rotational speed, friction is at its limiting value: \(\mu_s N = mg\). Step 3: Detailed Explanation:
Substitute the expression for Normal force into the limiting friction condition:
\[ \mu_s (m r \omega_{min}^2) = mg \]
The mass \(m\) cancels out from both sides:
\[ \mu_s r \omega_{min}^2 = g \]
Rearrange the equation to solve for the coefficient of friction \(\mu_s\):
\[ \mu_s = \frac{g}{r \omega_{min}^2} \]
Given values are \(g = 10 \text{ m/s}^2\), \(r = 4 \text{ m}\), and \(\omega_{min} = 5 \text{ rad/s}\).
Substitute these values into the formula:
\[ \mu_s = \frac{10}{4 \times (5)^2} \]
\[ \mu_s = \frac{10}{4 \times 25} = \frac{10}{100} = 0.1 \]
Step 4: Final Answer:
The coefficient of friction is \(0.1\).
