Question

For the following question, enter the correct numerical value upto TWO decimal places. If the numerical value has more than two decimal places, round-off the value to TWO decimal places. (For example: Numeric value 5 will be written as 5.00 and 2.346 will be written as 2.35) A body moves along a circular path of radius 10 m and the coefficient of friction is 0.5. What should be its angular velocity in rad/s if it is not to slip from the surface? ____. (\(g = 9.8 \, \text{m s}^{-2}\))

Hint:

For circular motion on a rough surface, friction provides the centripetal force. Always equate maximum frictional force to the required centripetal force to find limiting conditions.

Answer 1

Correct Answer: 0.7

Step 1: For a body moving in a horizontal circular path, the centripetal force required is provided by friction. Maximum frictional force: \[ F_f = \mu mg \]
Step 2: Centripetal force required: \[ F_c = m r \omega^2 \] For no slipping: \[ m r \omega^2 \leq \mu mg \]
Step 3: Cancel \(m\) from both sides: \[ r \omega^2 \leq \mu g \]
Step 4: Solve for angular velocity: \[ \omega^2 = \frac{\mu g}{r} \] \[ \omega = \sqrt{\frac{\mu g}{r}} \]
Step 5: Substitute the given values: \[ \omega = \sqrt{\frac{0.5 \times 9.8}{10}} = \sqrt{0.49} = 0.7 \, \text{rad/s} \] Final Answer (up to two decimal places): \[ \boxed{0.70} \]