A solid cylinder rolls down an inclined plane without slipping. If the translational kinetic energy of the cylinder is 140 J , the total kinetic energy of the cylinder is

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Aicharya · Accepted Answer

For solid cylinder, $\frac{\mathrm{K}^2}{\mathrm{R}^2}=\frac{1}{2}$ $\begin{aligned} & (\mathrm{K} \cdot \mathrm{E})_{	ext {Total }}=\frac{1}{2} \mathrm{~m} \mathrm{v}^2\left(1+\frac{\mathrm{K}^2}{\mathrm{R}^2}ight) \ & =(\mathrm{K} \cdot \mathrm{E})_{\mathrm{T}}\left(1+\frac{\mathrm{K}^2}{\mathrm{R}^2}ight)=140\left(1+\frac{1}{2}ight)=210 \mathrm{~J}\end{aligned}$

A solid cylinder rolls down an inclined plane without slipping. If the translational kinetic energy of the…

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