© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Understand the Given Condition
A body is rolling down an inclined plane. Its rotational kinetic energy (KR) is 40% (or 0.40 times) of its translational kinetic energy (KT). We need to identify the shape of this rolling object.
Step 2: Write Down the Relevant Equations
Translational kinetic energy:
$K_T = \frac{1}{2} M v^2.$
Rotational kinetic energy:
$K_R = \frac{1}{2} I \omega^2.$
Rolling condition (without slipping):
$\omega = \frac{v}{R}.$
Here, $M$ is the mass of the body, $I$ is its moment of inertia about the center of mass, $v$ is the linear speed, and $R$ is its radius.
Step 3: Express the Rotational Energy in Terms of v
From the rolling condition $\omega = \frac{v}{R}$, substitute into $K_R$:
$K_R = \frac{1}{2} I \left(\frac{v}{R}\right)^2
= \frac{1}{2} I \frac{v^2}{R^2}.$
Step 4: Apply the 40% Ratio
The given condition is $K_R = 0.40\,K_T$. Therefore,
$\frac{1}{2} I \frac{v^2}{R^2} = 0.40 \times \frac{1}{2} M v^2.$
Canceling common terms $\frac{1}{2} v^2$ from both sides gives:
$I \frac{1}{R^2} = 0.40\,M \quad
\Longrightarrow \quad I = 0.40\,M R^2.$
Since $0.40 = \frac{2}{5}$, we get:
$I = \frac{2}{5} M R^2.$
Step 5: Match the Moment of Inertia with Known Objects
Among standard rolling objects, the moment of inertia of a solid sphere (solid ball) about its center is $I_{\text{solid sphere}} = \frac{2}{5} M R^2.$
This exactly matches $I = \frac{2}{5} M R^2,$ which implies that for a solid ball rolling without slipping, its rotational kinetic energy is 40% of its translational kinetic energy.
Step 6: Conclusion
The body in question, satisfying $K_R = 0.40 \, K_T$, is therefore a solid ball.
