JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A uniform disc of mass \(0.5\,kg\) and radius \(r\) is projected with velocity \(18\,m / s\) at \(t =0\,s\) on a rough horizontal surface. It starts off with a purely sliding motion at \(t =0\,s\). After \(2\,s\) it acquires a purely rolling motion (see figure). The total kinetic energy of the disc after \(2\,s\) will be \(..............J\) (given, coefficient of friction is \(0.3\) and \(\left.g =10\,m / s ^2\right)\)

- A \(53\)
- B \(52\)
- C \(54\)
- D \(51\)
Answer & Solution
Correct Answer
(C) \(54\)
Step-by-step Solution
Detailed explanation
\(a =-\mu_{ k } g =-3\) \(V =18-3 \times 2\) \(V =12\,m / s\) \(KE =\frac{1}{2} mv ^2+\frac{1}{2} \frac{ mr ^2}{2} \frac{ v ^2}{ r ^2}\) \(KE =\frac{3}{4} mv ^2\) \(KE =3 \times 18=54\,J\)
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