System of Particles and Rotational Motion · Class 11 · CBSE · Science Physics

A thin uniform circular ring of mass M and radius R is rolling without slipping on a horizontal surface with a linear velocity v. The total kinetic e…

Practice question from System of Particles and Rotational Motion (Class 11 · CBSE · Science Physics), with the correct answer and a step-by-step solution.

easy · MCQ · Q-56588F

A thin uniform circular ring of mass $M$ and radius $R$ is rolling without slipping on a horizontal surface with a linear velocity $v$ . The total kinetic energy of the ring is:

A. $\frac{1}{2} Mv^2$
B. $Mv^2$ Correct answer
C. $\frac{3}{2} Mv^2$
D. $2 Mv^2$

Solution

Total kinetic energy for rolling without slipping is given by $K = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$ . For a ring, $I = MR^2$ and for rolling without slipping $\omega = v/R$ . Substituting these gives $K = \frac{1}{2}Mv^2 + \frac{1}{2}(MR^2)(v/R)^2 = \frac{1}{2}Mv^2 + \frac{1}{2}Mv^2 = Mv^2$ .

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