(A) For a solid sphere rolling down an inclined plane without slipping, the velocity $v$ at the bottom is given by the formula:$v = \sqrt{\frac{2gh}{1

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(A) For a solid sphere rolling down an inclined plane without slipping, the velocity $v$ at the bottom is given by the formula:$v = \sqrt{\frac{2gh}{1 + \frac{k^2}{R^2}}}$Where $g = 10 \text{ ms}^{-2}$, $h = 28 \text{ m}$, and for a solid sphere, the radius of gyration $k$ satisfies $k^2 = \frac{2}{5}R^2$, so $\frac{k^2}{R^2} = \frac{2}{5}$.Substituting the values:$v = \sqrt{\frac{2 \times 10 \times 28}{1 + \frac{2}{5}}}$$v = \sqrt{\frac{560}{\frac{7}{5}}}$$v = \sqrt{\frac{560 \times 5}{7}}$$v = \sqrt{80 \times 5} = \sqrt{400} = 20 \text{ ms}^{-1}$.

Class 11 Physics System of Particles and Rotational Motion Rolling On Inclined Plane

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$A$ solid sphere rolls down without slipping from the top of an inclined plane of height $28 \text{ m}$ and angle of inclination $30^{\circ}$. The velocity of the sphere, when it reaches the bottom of the plane is (Acceleration due to gravity $= 10 \text{ ms}^{-2}$)

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A
$20 \text{ ms}^{-1}$
B
$28 \text{ ms}^{-1}$
C
$10 \text{ ms}^{-1}$
D
$14 \text{ ms}^{-1}$

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System of Particles and Rotational Motion

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Rolling On Inclined Plane

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$A$ solid sphere rolls down without slipping from the top of an inclined plane of height $28 \text{ m}$ and angle of inclination $30^{\circ}$. The velocity of the sphere, when it reaches the bottom of the plane is (Acceleration due to gravity $= 10 \text{ ms}^{-2}$)

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