TS EAMCET · Physics · Laws of Motion
A child is on a merry-go-round, standing at a distance \(2 \mathrm{~m}\) from the centre. The coefficient of static friction between the child and the surface of merry-go-round is 0.8 . At what maximum angular velocity can the merry-go-round be rotated before the child slips? (Talse, \(g=10 \mathrm{~m} / \mathrm{s}^2\) )
- A \(0.5 \mathrm{rad} / \mathrm{s}\)
- B \(1 \mathrm{rad} / \mathrm{s}\)
- C \(2 \mathrm{rad} / \mathrm{s}\)
- D \(4 \mathrm{rad} / \mathrm{s}\)
Answer & Solution
Correct Answer
(C) \(2 \mathrm{rad} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
At critical speed, centrifugal force on child must be equal to maximum static friction \[ m \omega^2 r=\mu N \] where radius, \(r=2 \mathrm{~m}\), coefficient of static friction, \(\mu=0.8\) and \[ M=\text { mass of child. } \] Reaction force, \(N=m g\) So,…
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