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A thick walled hollow sphere has outer radius $\mathrm{R}$. It rolls down an inclined plain Without slipping and its speed at the bottom is v. If the inclined plane is frictionless and the sphere slides down without rolling, its speed at the bottom will be $5 \mathrm{v} / 4$. What is radius of gyration of the sphere.
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Verified Answer
The correct answer is:
$\frac{3 R}{4}$
Initially, when the sphere is rolling down the include without any slipping, then let the angular velocity of the sphere at the bottom of the incline be w.
No slipping condition, $\mathrm{V}=\mathrm{wR}$
Furthermore, due to conservation of energy
PE lost $=$ KE gained
$\mathrm{mgh}=1 / 2 \mathrm{mv}^{2}+1 / 2{ }^* \mathrm{I}^* \mathrm{w}^{2}$ (Where $\mathrm{h}$ is the height of the incline and $\mathrm{I}$ is the moment of inertia of sphere around its center)
Also, $\mathrm{I}=\mathrm{m}^* \mathrm{Rg}^{2}$ where $\mathrm{Rg}$ is the radius of gyration and also $\mathrm{v}=\mathrm{wR}$ (No slipping condition)
Hence, $\mathrm{mgh}=1 / 2 \mathrm{mv}^{2}+1 / 2{ }^* \mathrm{~m} * \mathrm{Rg}{ }^{2}{ }^* \mathrm{v}^{2} / \mathrm{R}^{2}$ $2 \mathrm{gh}=\left[1+(\mathrm{Rg} / \mathrm{R})^{2}\right] * \mathrm{v}^{2}(1)$
Furthermore, under no rolling case
$\begin{aligned}& \mathrm{mgh}=1 / 2 \mathrm{~m}^*(5 \mathrm{v} / 4)^{2} \\& \mathrm{v}^{2}=32 / 25 \mathrm{gh}\end{aligned}$
Putting it in the equation 1
$2 \mathrm{gh}=\left[1+(\mathrm{Rg} / \mathrm{R})^{2}\right]^* 32 / 25 \mathrm{gh}$
Hence, $\mathrm{Rg}=3 / 4 \mathrm{R}$
So, the radius of gyration of the sphere around its center is $3 / 4 \mathrm{R}$
No slipping condition, $\mathrm{V}=\mathrm{wR}$
Furthermore, due to conservation of energy
PE lost $=$ KE gained
$\mathrm{mgh}=1 / 2 \mathrm{mv}^{2}+1 / 2{ }^* \mathrm{I}^* \mathrm{w}^{2}$ (Where $\mathrm{h}$ is the height of the incline and $\mathrm{I}$ is the moment of inertia of sphere around its center)
Also, $\mathrm{I}=\mathrm{m}^* \mathrm{Rg}^{2}$ where $\mathrm{Rg}$ is the radius of gyration and also $\mathrm{v}=\mathrm{wR}$ (No slipping condition)
Hence, $\mathrm{mgh}=1 / 2 \mathrm{mv}^{2}+1 / 2{ }^* \mathrm{~m} * \mathrm{Rg}{ }^{2}{ }^* \mathrm{v}^{2} / \mathrm{R}^{2}$ $2 \mathrm{gh}=\left[1+(\mathrm{Rg} / \mathrm{R})^{2}\right] * \mathrm{v}^{2}(1)$
Furthermore, under no rolling case
$\begin{aligned}& \mathrm{mgh}=1 / 2 \mathrm{~m}^*(5 \mathrm{v} / 4)^{2} \\& \mathrm{v}^{2}=32 / 25 \mathrm{gh}\end{aligned}$
Putting it in the equation 1
$2 \mathrm{gh}=\left[1+(\mathrm{Rg} / \mathrm{R})^{2}\right]^* 32 / 25 \mathrm{gh}$
Hence, $\mathrm{Rg}=3 / 4 \mathrm{R}$
So, the radius of gyration of the sphere around its center is $3 / 4 \mathrm{R}$
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