Question

A solid ball of mass m=2kg is in pure rolling over a horizontal surface with velocity of centre of mass v=10m/s the kinetic energy of the ball is

Answer 1

Explanation:

given,

m=2

v=10

KE=1/2mv square

KE=1/2x2x100

KE=100ms

Answer 2

Answer:

The kinetic energy of the ball is 140 J

Explanation:

Given,

mass of the object (m) = 2 Kg and velocity of the centre of mass (v) = 10 m/s

to find,

The kinetic energy of the ball(K.E)

We know that kinetic energy of the ball(K.E) is the combined kinetic energy of translation motion ($K.E_T$) and of the circular motion ($K.E_R$)

i.e.

$K.E = K.E_T +K.E_R$

$K.E = \frac{1}{2} mv^2 + \frac{1}{2} I\omega^2$  (where I = moment of inertia and ω = angular velocity = v/r)

$K.E = \frac{1}{2} mv^2 + \frac{1}{2} (\frac{2}{5}mr^2 )(\frac{v}{r} )^2$ (where I = 2/5 mr²)

$K.E = \frac{1}{2} mv^2 + \frac{1}{2} \frac{2}{5}mv^2$

$K.E. = \frac{7}{10} mv^2$

$K.E = \frac{7}{10} (2)(10)^2$

K.E = 140 J

Therefore,  the kinetic energy of the ball is 140 J.

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