A solid sphere of radius r is revolving about one of its diameters with an angular velocity ω. If it suddenly expands uniformly so that its radius increases to n times its original value, then its angular velocity becomes

Question

A solid sphere of radius r is revolving about one of its diameters with an angular velocity ω. If it suddenly expands uniformly so that its radius increases to n times its original value, then its angular velocity becomes (A) n2 ω (B) (ω/n2) (C) n ω (D) (ω/n)

Tardigrade Admin · Accepted Answer

I1 ω1=I2 ω2 (2/5) m r2 ω=(2/5) m(r')2 × ω2 r2 ω=(n r)2 ω2 r2 ω=n2 r2 ω2 ω2=(ω/n2)

Q. A solid sphere of radius $r$ is revolving about one of its diameters with an angular velocity $ω$ . If it suddenly expands uniformly so that its radius increases to $n$ times its original value, then its angular velocity becomes

$n^{2} ω$

$\frac{ω}{n ^{2}}$

$nω$

$\frac{ω}{n}$

$2 nω$

$I_{1} ω_{1} = I_{2} ω_{2}$
$\frac{2}{5} m r^{2} ω = \frac{2}{5} m (r^{'})^{2} \times ω_{2}$
$r^{2} ω = (n r)^{2} ω_{2}$
$r^{2} ω = n^{2} r^{2} ω_{2}$
$ω_{2} = \frac{ω}{n ^{2}}$

Q. A solid sphere of radius r is revolving about one of its diameters with an angular velocity ω. If it suddenly expands uniformly so that its radius increases to n times its original value, then its angular velocity becomes

n2ω

n2ω​

nω

nω​

2nω

I1​ω1​=I2​ω2​ 52​mr2ω=52​m(r′)2×ω2​ r2ω=(nr)2ω2​ r2ω=n2r2ω2​ ω2​=n2ω​