A copper ball of mass `m = 1 kg` with a radius of `r = 10 cm` rotates with angular velocity `omega = 2 rad//s` about an axis passing through its centre. The work should be performed to increase the angular velocity of rotation of the ball two fold is.

Options

A. `1.2 xx 10^-2 J`
B. `2.4 xx 10^-2 J`
C. `3.6 xx 10^-2 J`
D. `4.8 xx 10^-2 J`

Answer

B

Explanation

(b) `K.E. = (1)/(2) ((2)/(5)) mv^2 rArr K.E. = (2)/(10) mv^2`
If `omega` is to be doubled, `v` is also doubled.
Therefore, `Delta K.E. = K.E._2 - K.E._1 = (2)/(10) m[v_2^2 - v_1^2]`
`v= 10 - t = 0 Delta K.E. = (2)/(10) m[r^2 omega_2^2 - r^2 omega_1^2]`
=`(2)/(10) xx 1[((1)/(10))^2][4^2 - 2^2] = 2.4 xx 10^-2 J`.