JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A disc of radius \(R\) and mass \(M\) is rolling horizontally without slipping with speed \(v\). It then moves up an inclined smooth surface as shown in figure. The maximum height that the disc can go up the incline is _______.
- A \(\frac{v^2}{g}\)
- B \(\frac{3}{4} \frac{v^2}{g}\)
- C \(\frac{1}{2} \frac{v^2}{g}\)
- D \(\frac{2}{3} \frac{v^2}{g}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{2} \frac{v^2}{g}\)
Step-by-step Solution
Detailed explanation
Only the translational kinetic energy of disc changes into gravitational potential energy. And rotational \(\mathrm{KE}\) remains unchanged as there is no friction. \(\frac{1}{2} \mathrm{mv}^2=\mathrm{mgh}\) \(\mathrm{h}=\frac{\mathrm{v}^2}{2 \mathrm{~g}}\)
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