A solid sphere and a disc of same mass M and | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The moment of inertia of a solid sphere about an axis passing through its center is $I_{cm, sphere} = \frac{2}{5} M R^2$. Using the parallel axis

Vedclass · Accepted Answer

$\frac{29 M R^2}{10}$

Vedclass · Answer

(C) The moment of inertia of a solid sphere about an axis passing through its center is $I_{cm, sphere} = \frac{2}{5} M R^2$. Using the parallel axis theorem, the moment of inertia about an axis passing through its edge (point of contact) is $I_1 = I_{cm, sphere} + M R^2 = \frac{2}{5} M R^2 + M R^2 = \frac{7}{5} M R^2$.The moment of inertia of a disc about an axis passing through its center and perpendicular to its plane is $I_{cm, disc} = \frac{1}{2} M R^2$. Using the parallel axis theorem, the moment of inertia about an axis passing through its edge (point of contact) is $I_2 = I_{cm, disc} + M R^2 = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2$.The total moment of inertia of the system about the axis passing through the point of contact is $I = I_1 + I_2 = \frac{7}{5} M R^2 + \frac{3}{2} M R^2 = \frac{14 + 15}{10} M R^2 = \frac{29}{10} M R^2$.

Similar Questions

$\frac{1+\tanh \frac{x}{2}}{1-\tanh \frac{x}{2}}$ is equal to

In an atom,the order of increasing energy of electrons with quantum numbers $(i)$ $n=4, l=1$,$(ii)$ $n=4, l=0$,$(iii)$ $n=3, l=2$,and $(iv)$ $n=3, l=1$ is:

$\int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}} \, dx = $

In which of the following reactions,$H_2O_2$ acts as a reducing reagent?

The wavelengths of two sound notes in air are $\frac{40}{195} ~m$ and $\frac{40}{193} ~m$. Each note produces $9$ beats per second separately with a third note of fixed frequency. The velocity of sound in air in $m/s$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module