Question 416: Centre of Mass and Rotational Motion

Question

A solid sphere and a solid cylinder of same mass and radius are rolling on a horizontal surface without slipping. The ratio of their radius of gyrations respectively $\left(k_{	ext {sph }}: k_{	ext {cyl }}ight)$ is $2: \sqrt{x}$. The value of $x$ is ____________ .

Accepted Answer

Let the mass of both the solid sphere and the solid cylinder be $M$, and let their common radius be $R$. The moment of inertia $I$ of a solid sphere and a solid cylinder are given by:
$I_{sph} = \frac{2}{5}MR^2$
$I_{cyl} = \frac{1}{2}MR^2$
The radius of gyration $k$ is related to the moment of inertia $I$ by the formula $I = Mk^2$. Therefore, we can find the radius of gyration for both the solid sphere and the solid cylinder using their respective moments of inertia:
$k_{sph}^2 = \frac{I_{sph}}{M} = \frac{2}{5}R^2$
$k_{cyl}^2 = \frac{I_{cyl}}{M} = \frac{1}{2}R^2$
Now, let's find the ratio of their radius of gyrations:
$\frac{k_{sph}}{k_{cyl}} = \frac{2}{\sqrt{x}}$
Squaring both sides:
$\frac{k_{sph}^2}{k_{cyl}^2} = \frac{4}{x}$
Substituting the expressions for $k_{sph}^2$ and $k_{cyl}^2$:
$\frac{\frac{2}{5}R^2}{\frac{1}{2}R^2} = \frac{4}{x}$
Simplifying and solving for $x$:
$\frac{2}{5} \cdot \frac{2}{1} = \frac{4}{x}$
$\frac{4}{5} = \frac{4}{x}$
Thus, $x = 5$.

Centre of Mass and Rotational Motion

A solid sphere and a solid cylinder of same mass and radius are rolling on a horizontal surface without slipping. The ratio of their radius of gyrations respectively \left(k_{\text {sph }}: k_{\text {cyl }}\right) is 2: \sqrt{x}. The value of x is ____________ .

Integer / Numeric Answer Type