Question

A thin hollow sphere of mass \[m\] is completely filled with a liquid of mass \[m\] . When the sphere rolls with a velocity \[v\] , kinetic energy of the system is (neglect friction):
A. $\dfrac{1}{2}m{v^2}$
B. $m{v^2}$
C. $\dfrac{4}{3}m{v^2}$
D. $\dfrac{4}{5}m{v^2}$

Answer 1

Hint:To answer the given question, we will use the concept of total kinetic energy of a system, which is the sum of kinetic energy and rotational kinetic energy as the sphere rolls, to find the kinetic energy of the system of a thin hollow sphere filled with liquid of mass \[m\] and rolling with a velocity \[v\].

Complete step by step answer:
The sum of the kinetic energies originating from each type of motion equals the total kinetic energy of a body or system.Now, let us coming to the given question, as we discussed in concept.Kinetic energy of rotation is given by –
$K = \dfrac{1}{2}I{\omega ^2} \\
\Rightarrow K= \dfrac{1}{2} \times \dfrac{2}{3}m{v^2}\dfrac{{{v^2}}}{{{r^2}}} \\
\Rightarrow K= \dfrac{1}{3}m{v^2}$
Where, $I$ is the moment of inertia about axis of rotation and $\omega $ is the angular velocity and the value of $I$ for sphere is $\dfrac{2}{3}m{r^2}$.

And kinetic energy of liquid is given by-
$\dfrac{1}{2}m{v^2} + \dfrac{1}{2}m{v^2} = m{v^2}$
Therefore, the kinetic energy of the system will be
$\text{Total energy} = K.E + \text{rotational K.E}$
$\Rightarrow \text{Total energy}= m{v^2} + \dfrac{1}{3}m{v^2} \\
\therefore \text{Total energy}= \dfrac{4}{3}m{v^2} \\ $
Therefore, kinetic energy of the system is $\dfrac{4}{3}m{v^2}$.

Therefore, the correct option is C.

Note: It’s worth mentioning that a rotating rigid body's kinetic energy is exactly proportional to its moment of inertia and the square of its angular velocity. Flywheel energy storage devices, which are designed to store enormous amounts of rotational kinetic energy, take use of this. Many carmakers are currently putting flywheel energy storage systems, such as the flywheel or kinetic energy recovery system, to the test in their vehicles.