6.System of Particles and Rotational Motion
medium

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 .......... $J$

A

$50$

B

$100$

C

$150$

D

$175$

Solution

(b)

Using mechanical energy conservation

$m g h=\frac{1}{2} m v^2+\frac{1}{2} l \omega^2$

$3(5)(10)=\frac{1}{2} m v^2+\frac{1}{2} m l^2\left(\frac{v^2}{l^2}\right)$

$150=\frac{3}{4} m v^2$

$m v^2=200$

$\frac{1}{2} m v^2=100 J = K \cdot E _{\cdot} \cdot \text { Translation }$

Standard 11
Physics

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