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Question

A solid cylinder with a radius of 4.0 cm has the same mass as a solid sphere of radius R. If the cylinder and sphere have the same moment of inertia about their centers, what is the sphere's radius?

Solution

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Given values:\text{\underline{\textbf{Given values:}}}

2r=4cm2r=4 \: \text{cm}

r=2cm=0.02mr=2\: \text{cm}=0.02 \: \text{m}

The moment of inertia of the cylinder is equal to:

I1=12m1r12\begin{align*} I_1=\frac{1}{2}m_1{r_1}^2 \tag{Equation $1.$} \end{align*}

The moment of inertia of the sphere is equal:

I2=25m2r22\begin{align*} I_2=\frac{2}{5}m_2{r_2}^2 \tag{Equation $2.$} \end{align*}

As we know that the moments of the cylinder and the sphere are equal, the mass of the cylinder and the sphere is also equal by equalizing equations 11 and 22, we find the diameter of the sphere:

I1=I2m1=m2I=12mr12I=25mr2212mr12=25mr2254r12=r22r2=r1254r2=r125r2=0.02236067m    2r2=0.0447m\begin{align*} I_1&=I_2\\ m_1&=m_2\\ I&=\frac{1}{2}m{r_1}^2\\ I&=\frac{2}{5}m{r_2}^2\\ \frac{1}{2}m{r_1}^2&=\frac{2}{5}m{r_2}^2\\ \frac{5}{4}{r_1}^2&={r_2}^2\\ r_2&=\sqrt {{r_1}^2 \frac{5}{4} }\\ r_2&=\frac{r_1}{2}\sqrt {5}\\ r_2&=0.02236067 \: \text{m}\\\\ \implies 2r_2&=\boxed{0.0447 \: \text{m}}\\ \end{align*}

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