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Home » A ring and a disc roll on the horizontal surface without slipping with same linear velocity. If both have same mass and total kinetic energy of the ring is then total kinetic energy of the disc isA)3 JB)4 JC)5 JD)6 J
A ring and a disc roll on the horizontal surface without slipping with same linear velocity. If both have same mass and total kinetic energy of the ring is 4 \mathrm{~J} then total kinetic energy of the disc is
A)3 J
B)4 J
C)5 J
D)6 J
Physics
A ring and a disc roll on the horizontal surface without slipping with same linear velocity. If both have same mass and total kinetic energy of the ring is 4 \mathrm{~J} then total kinetic energy of the disc is
A)3 J
B)4 J
C)5 J
D)6 J

Correct option is A.

Total kinetic energy of the body \mathrm{K} \cdot \mathrm{E}_{\text {total }}=\left(\frac{1}{2} \mathrm{mv}^{2}\right)_{\text {translational }}+ \left(\frac{1}{2} \mathrm{Iw}^{2}\right)_{\text {rotational }}

Total kinetic energy of the ring K, E_{\text {ring }}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2}\left(\mathrm{Mr}^{2}\right) \mathrm{w}^{2}

\begin{array}{l} \text { K. } \left.E_{\text {ring }}=\frac{1}{2} m v^{2}+\frac{1}{2} M v^{2}=M v^{2} \quad \text { (for pure rolling } v=r w\right) \\ \Longrightarrow m v^{2}=4 \end{array}

Total kinetic energy of the ring \mathrm{K}, \mathrm{E}_{\text {ring }}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2}\left(\frac{1}{2} \mathrm{Mr}^{2}\right) \mathrm{w}^{2}
\begin{array}{l} \left.K \cdot E_{\text {disc }}=\frac{1}{2} m_{v}^{2}+\frac{1}{4} M v^{2}=\frac{3}{4} \mathrm{Mv}^{2} \quad \text { (for pure rolling } \mathrm{v}=\mathrm{rw}\right) \\ \Longrightarrow \mathrm{K}, \mathrm{E}_{\text {disc }}=\frac{3}{4} \times 4=3 \mathrm{~J} \end{array}

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