Correct Answer - Option 4 : 7 : 2
The correct answer is option 4) i.e. 7 : 2
CONCEPT:
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Rotational kinetic energy: Rotational kinetic energy is due to the motion of an object's rotation.
For a given fixed axis of rotation, the rotational kinetic energy is given by:
\(KE = \frac{1}{2} Iω^2\)
Where I is the moment of inertia, ω is the angular velocity.
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Moment of inertia of a sphere about its central axis:
- For a sphere of mass M and radius R, the moment of inertia (IC) about an axis passing through its centre is given by
\(I_C = {\frac{2}{5}}MR^2\)
CALCULATION:
- The rolling motion is a combination of translation and rotation.
- The rolling kinetic energy of a sphere is the sum of the translational kinetic energy of its centre of mass plus the rotational kinetic energy about the centre of mass.
- The translational kinetic energy of an object of mass m and velocity v is given as \( K_1 = \frac{1}{2}mv^2\)
- Rotational kinetic energy, \(K_2 = \frac{1}{2} Iω^2\)
- Rolling kinetic energy, K3 = K1 + K2
Ratio = \(\frac{K_3}{K_2} = \frac{K_1 + K_2}{K_2} = \frac{K_1}{K_2} + 1\)
On substituting the values,
\(\Rightarrow \frac{K_3}{K_2} = \frac{\frac{1}{2}mv^2}{\frac{1}{2}Iω^2} + 1\)
\(\Rightarrow \frac{K_3}{K_2} = \frac{mv^2}{Iω^2} + 1\)
We know that linear velocity, v = rω and moment of inertia for a sphere, \(I = {\frac{2}{5}}MR^2\)
\(\Rightarrow \frac{K_3}{K_2} = \frac{m(r\omega)^2}{\frac{2}{5}mr^2ω^2} + 1=\frac{5}{2} +1 \)
\(\Rightarrow \frac{K_3}{K_2} = \frac{7}{2}\)
