TS EAMCET · Physics · Laws of Motion
A uniform cylinder of radius \(1 \mathrm{~m}\), mass \(1 \mathrm{~kg}\) spins about its axis with an angular velocity \(20 \mathrm{rad} / \mathrm{s}\). At certain moment, the cylinder is placed into a corner as shown in the figure. The coefficient of friction between the horizontal wall and the cylinder is \(\mu\), whereas the vertical wall is frictionless. If the number of rounds made by the cylinder is 5 before it stops, then the value of \(\mu\) is (acceleration due to gravity, \(g=10 \mathrm{~m} / \mathrm{s}^2\) ) 
- A \(\frac{3}{\pi}\)
- B \(\frac{2}{\pi}\)
- C \(\frac{1}{\pi}\)
- D \(\frac{0.4}{\pi}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{\pi}\)
Step-by-step Solution
Detailed explanation
Free body diagram of cylinder is We have following equations, \[ \begin{aligned} N_2 & =m g \\ \mu N_2 & =N_1 \end{aligned} \] Torque about centre of cylinder is \[ \mu N_2 R=\frac{m R^2}{2} \alpha \] From Eqs. (i) and (iii), we get…
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