A ring of radius 2 m weights 100 kg . It is having pure rolling motion on a horizontal floor so that its centre of mass has a speed of 20 cms- 1 . If work done to stop it is xJ . Then x will be

Question

A ring of radius 2 m weights 100 kg . It is having pure rolling motion on a horizontal floor so that its centre of mass has a speed of 20 cms- 1 . If work done to stop it is xJ . Then x will be (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

W=Δ K =(1/2)(mv)2(1 + (K2/R2)) =(1/2)(mv)2(.1+1.) =100(0.2 =4 J

Q. A ring of radius $2 m$ weights $100 k g$ . It is having pure rolling motion on a horizontal floor so that its centre of mass has a speed of $20 c m s^{- 1}$ . If work done to stop it is $x J$ . Then $x$ will be

$W = Δ K$
$= \frac{1}{2} (m v)^{2} (1 + \frac{K ^{2}}{R ^{2}})$
$= \frac{1}{2} (m v)^{2} (1 + 1)$
$= 100 (0.2$
$= 4 J$

Q. A ring of radius 2m weights 100kg . It is having pure rolling motion on a horizontal floor so that its centre of mass has a speed of 20cms−1 . If work done to stop it is xJ . Then x will be

W=ΔK =21​(mv)2(1+R2K2​) =21​(mv)2(1+1) =100(0.2 =4J

Q. A ring of radius $2 m$ weights $100 k g$ . It is having pure rolling motion on a horizontal floor so that its centre of mass has a speed of $20 c m s^{- 1}$ . If work done to stop it is $x J$ . Then $x$ will be

$W = Δ K$
$= \frac{1}{2} (m v)^{2} (1 + \frac{K ^{2}}{R ^{2}})$
$= \frac{1}{2} (m v)^{2} (1 + 1)$
$= 100 (0.2$
$= 4 J$