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         a coin is kept a distance of 10cms from the centre of a turnable table .If the coefficient of  static friction is 0.8 find the frequency of rotation of the disc at which the coin will begin to slip
2 years ago

Pushkar Prateek
23 Points
							Suppose the turn table rotates with  angular (frequency or speed) $\omega$ rad/s. Here, friction will provide the required centripetal force.The coin will begin to slip when required centripetal force just exceeds maximum possible friction.So in limiting conditionCentripetal force required for circular motion of the coin just becomes equal to maximum possible friction. 0.8mg = m $\omega$2 0.1$\omega$ = 4$\sqrt{5}$ rad/s.

2 years ago
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• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions