COMEDK · Physics · 5. Laws of Motion
A coin is placed on a disc rotating with an angular velocity \(\omega\). The co-efficient of friction between the disc and the coin is \(\mu\). The maximum distance of the coin from the centre of the disc up to which it will rotate with the disc is
- A \(\dfrac{\mu g}{\omega^2}\)
- B \(\sqrt{\dfrac{\mu g}{\omega^2}}\)
- C \(\dfrac{\mu g}{\omega}\)
- D \(\sqrt{\dfrac{\mu}{\omega^2}}\)
Answer & Solution
Correct Answer
(A) \(\dfrac{\mu g}{\omega^2}\)
Step-by-step Solution
Detailed explanation
For the coin to rotate with the disc without slipping, the centripetal force required for circular motion must be provided by the static frictional force. Let \(m\) be the mass of the coin, \(r\) be the distance from the centre, and \(\omega\) be the angular velocity. The…
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