A spherical shell of 1 kg mass and radius R is rolling with angular speed ω on horizontal plane (as shown in figure). The magnitude of angular momentum of the shell about the origin O is (a/3) R2 ω. The value of a will be :

Question

A spherical shell of 1 kg mass and radius R is rolling with angular speed ω on horizontal plane (as shown in figure). The magnitude of angular momentum of the shell about the origin O is (a/3) R2 ω. The value of a will be : (A) 2 (B) 3 (C) 5 (D) 4

Tardigrade Admin · Accepted Answer

L 0= angular momentum of shell about O. As shell is rolling so V cm =ω R L 0= mV cm R + I ω =1 × ω R × R +(2/3) R 2 ω =(5/3) R 2 ω so a =5

Q. A spherical shell of $1 k g$ mass and radius $R$ is rolling with angular speed $ω$ on horizontal plane (as shown in figure). The magnitude of angular momentum of the shell about the origin $O$ is $\frac{a}{3} R^{2} ω$ . The value of a will be :

2

3

5

4

$L_{0} =$ angular momentum of shell about $O$ .
As shell is rolling
so $V_{c m} = ω R$
$L_{0} = m V_{c m} R + I ω$
$= 1 \times ω R \times R + \frac{2}{3} R^{2} ω$
$= \frac{5}{3} R^{2} ω$
so $a = 5$

Q. A spherical shell of 1kg mass and radius R is rolling with angular speed ω on horizontal plane (as shown in figure). The magnitude of angular momentum of the shell about the origin O is 3a​R2ω. The value of a will be :

2

3

5

4

L0​= angular momentum of shell about O. As shell is rolling so Vcm​=ωR L0​=mVcm​R+Iω =1×ωR×R+32​R2ω =35​R2ω so a=5

Q. A spherical shell of $1 k g$ mass and radius $R$ is rolling with angular speed $ω$ on horizontal plane (as shown in figure). The magnitude of angular momentum of the shell about the origin $O$ is $\frac{a}{3} R^{2} ω$ . The value of a will be :

$L_{0} =$ angular momentum of shell about $O$ .
As shell is rolling
so $V_{c m} = ω R$
$L_{0} = m V_{c m} R + I ω$
$= 1 \times ω R \times R + \frac{2}{3} R^{2} ω$
$= \frac{5}{3} R^{2} ω$
so $a = 5$