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  4. A disc of mass 3 kg rolls down an inclined plane of height 5 m. The translational kinetic energy of the disc on reaching the bottom of the inclined plane is

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Q. A disc of mass $3\, kg$ rolls down an inclined plane of height $5\, m$. The translational kinetic energy of the disc on reaching the bottom of the inclined plane is

System of Particles and Rotational Motion

Solution:

Using mechanical energy conservation
$m g h=\frac{1}{2} m v^{2}+\frac{1}{2} l \omega^{2}$
$3(5)(10)=\frac{1}{2} m v^{2}+\frac{1}{2} m l^{2}\left(\frac{v^{2}}{l^{2}}\right)$
$150=\frac{3}{4} m v^{2}$
$m v^{2}=200$
$\frac{1}{2} m v^{2}=100\, J = K . E._{\text{Translation}}$