Question

solid sphere is rotating about a diameter at an angular velocity w. if it cools so that its radius reduces to 1/n of its original value. It's time period of rotation is

Answer 1

Answer:

...........................................................??.............

Answer 2

Answer:

w.n^2

Explanation:

for solid sphere, MI = (2/5)MR2

angular momentum is always conserved.

L1 = L2

I1ω1 = I2ω2

here ω1 = ω

hence

ω2 = [(I1ω) / I2]

(1) Initial MI = I1 = (2/5)MR12

Radius is reduced to (1/n) of original value hence

R2 = (R1 / n)

hence

new MI = I2 = (2/5)MR22

from (1)

ω2 = (ω)[{(2/5)MR12} / {(2/5)MR22}] = (ω)(R1 / R2)2

ω2 = ω ∙ n^2 from (2)