For a rolling spherical shell, the ratio of rotational kinetic energy and total kinetic energy is $\frac{x}{5}$. The value of $x$ is ___________.
Answer (integer)
2
Solution
For a rolling spherical shell, we must consider the fact that it has both translational and rotational kinetic energy. The total kinetic energy ($K_{total}$) can be expressed as the sum of the translational kinetic energy ($K_{trans}$) and the rotational kinetic energy ($K_{rot}$):
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$K_{total} = K_{trans} + K_{rot}$
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The translational kinetic energy of an object with mass (m) and linear velocity (v) is given by:
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$K_{trans} = \frac{1}{2}mv^2$
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The rotational kinetic energy of a rolling spherical shell with moment of inertia (I) and angular velocity (ω) is given by:
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$K_{rot} = \frac{1}{2}Iω^2$
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For a rolling object without slipping, the relationship between linear velocity (v) and angular velocity (ω) is:
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$v = Rω$
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Where R is the radius of the spherical shell.
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The moment of inertia for a spherical shell is given by:
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$I = \frac{2}{3}mR^2$
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Now, we can substitute the moment of inertia and the relationship between linear and angular velocity into the equation for rotational kinetic energy:
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$K_{rot} = \frac{1}{2}\left(\frac{2}{3}mR^2\right)\left(\frac{v}{R}\right)^2$
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Simplifying the equation:
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$K_{rot} = \frac{1}{2}\left(\frac{2}{3}mR^2\right)\frac{v^2}{R^2}$<br/><br/>
$K_{rot} = \frac{1}{3}mv^2$
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Now, we can find the ratio of rotational kinetic energy to total kinetic energy:
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$$\frac{K_{rot}}{K_{total}} = \frac{\frac{1}{3}mv^2}{\frac{1}{2}mv^2 + \frac{1}{3}mv^2}$$
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Simplifying the equation:
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$$\frac{K_{rot}}{K_{total}} = \frac{\frac{1}{3}}{\frac{1}{2} + \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{5}{6}}$$
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Multiplying both the numerator and the denominator by 6:
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$\frac{K_{rot}}{K_{total}} = \frac{2}{5}$
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Comparing this to the given ratio of $\frac{x}{5}$, we can determine that the value of $x$ is 2.
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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