Question

A solid cylinder of mass M and radius R rolls down an inclined plane of height h without slipping. The speed of its centre when it reaches the bottom is.

Answer 1

Answer:

Work-energy theorem,        Mgh=21Mv2+21Iw2

For solid cylinder,  I=21MR2

Also   v=Rw       (due to pure rolling)

Thus     Mgh=21Mv2+21×21MR2w2

$$\implies  v= \sqrt{\dfrac{4}{3}gh}$$

Answer 2

Answer:

correct answer

$\sqrt{(4 \div 3)gh}$

Explanation:

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