A hollow sphere of mass m filled with a non-v | vedclass

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

Question

(A) The total mass of the system is $M = m + m = 2m$. Since the liquid is non-viscous and the inner surface is frictionless,the liquid does not rotate

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) The total mass of the system is $M = m + m = 2m$. Since the liquid is non-viscous and the inner surface is frictionless,the liquid does not rotate with the sphere. Thus,the moment of inertia of the system about the center of mass is only due to the hollow sphere,$I = \frac{2}{3}mR^2$.For pure rolling,$v = \omega R$,so $\omega = v/R$.The translational kinetic energy is $K_{trans} = \frac{1}{2} M v^2 = \frac{1}{2} (2m) v^2 = mv^2$.The rotational kinetic energy is $K_{rot} = \frac{1}{2} I \omega^2 = \frac{1}{2} (\frac{2}{3} mR^2) (v/R)^2 = \frac{1}{3} mv^2$.The ratio is $\frac{K_{trans}}{K_{rot}} = \frac{mv^2}{\frac{1}{3} mv^2} = \frac{3}{1}$.Thus,$x = 3$ and $y = 1$. The value of $(x+y) = 3+1 = 4$.

Similar Questions

$A$ uniform spherical object of mass $M$ and radius $R$ has a moment of inertia $I$. It rolls without slipping down an inclined plane of angle $\theta$. What is its acceleration?

$A$ sphere rolls down on an inclined plane of inclination $\theta$. What is the acceleration as the sphere reaches the bottom?

$A$ hollow sphere of mass $m$ and radius $R$ is rolling downward on a rough inclined plane of inclination $\theta$. If the coefficient of friction between the hollow sphere and incline is $\mu$,then ........

$A$ sphere,a disc,a ring,and a hollow sphere of the same radius are rolled down from the same height on an inclined plane simultaneously. The order in which these objects reach the bottom is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module