A solid sphere of mass 2 kg is making pure rolling on a horizontal surface with kinetic energy 2240 J. The velocity of centre of mass of the sphere will be ms -1.

Question

A solid sphere of mass 2 kg is making pure rolling on a horizontal surface with kinetic energy 2240 J. The velocity of centre of mass of the sphere will be ms -1. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

KE =(1/2) mv 2+(1/2) I ω2 2240=(1/2) 2( v )2+(1/2) (2/5)(2) R 2 ⋅(( v / R ))2 2240= v 2+(2/5) v 2 ⇒ v =40 m / s

Q. A solid sphere of mass $2 k g$ is making pure rolling on a horizontal surface with kinetic energy $2240 J$ . The velocity of centre of mass of the sphere will be ____ $m s^{- 1}$ .

$K E = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$
$2240 = \frac{1}{2} 2 (v)^{2} + \frac{1}{2} \frac{2}{5} (2) R^{2} \cdot (\frac{v}{R})^{2}$
$2240 = v^{2} + \frac{2}{5} v^{2}$
$\Rightarrow v = 40 m / s$

Q. A solid sphere of mass 2kg is making pure rolling on a horizontal surface with kinetic energy 2240J. The velocity of centre of mass of the sphere will be ____ms−1.

KE=21​mv2+21​Iω2 2240=21​2(v)2+21​52​(2)R2⋅(Rv​)2 2240=v2+52​v2 ⇒v=40m/s

Q. A solid sphere of mass $2 k g$ is making pure rolling on a horizontal surface with kinetic energy $2240 J$ . The velocity of centre of mass of the sphere will be ____ $m s^{- 1}$ .

$K E = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$
$2240 = \frac{1}{2} 2 (v)^{2} + \frac{1}{2} \frac{2}{5} (2) R^{2} \cdot (\frac{v}{R})^{2}$
$2240 = v^{2} + \frac{2}{5} v^{2}$
$\Rightarrow v = 40 m / s$