A solid cylinder of mass m and radius R rolls down an inclined

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

K.E. =	1	Iω² +	1	mv²
	2		2

K.E. =	1		1	mr²		ω² +	1	mv²
	2		2				2

=	1	mv² +	1	mv² =	3	mv²
	4		2		4

Now, gain in K.E. = Loss in P.E.

3	mv² = mgh ⇒ v - √		4		gh
4			3