Question

A) About which axis of rotation, radius of gyration is the least? B) State the formula for the moment of inertia of a solid sphere and hollow sphere about its diameter. C) Find the ratio of radius of gyration of a solid sphere about its diameter to radius of gyration of hollow sphere about its tangent. (Given-radius of both the spheres is the same). D) State SI unit and dimensions of angular momentum of rotating body.

Answer 1

Hint: First we need to know what is the radius of gyration of a body. Then we will determine how it can be minimum. For the next part we also need to know the formulae for the moment of inertia about different axes for a solid and hollow sphere. For the last part we need to know the definition of the angular momentum of a rotating body.
Formula used:
$I_t=I+M{R}^2$

Complete answer:
A) The radius of gyration of a body about an axis is defined as the radial distance of a point from the axis where the whole mass of the body may be assumed to be concentrated so that the moment of inertia about the axis is same with that of the actual distribution of mass of the body. Clearly if the axis of rotation passes through the centre of mass of the body, the radius of gyration would be minimum.

B) The formulae for moment of inertia of a solid sphere and a hollow sphere about its diameter is given by
$I={\displaystyle \frac{2}{5}}M{R}^2$ and $I={\displaystyle \frac{2}{3}}M{R}^2$ respectively, where $M$ and$R$ are mass and radius of the spheres respectively.

C)Now the moment of inertia of a hollow sphere about a tangent, we apply parallel axes theorem and it is given by
$\begin{array}{rl}& I_t=I+M{R}^2\\ & ={\displaystyle \frac{2}{3}}M{R}^2+M{R}^2\\ & ={\displaystyle \frac{5}{3}}M{R}^2\end{array}$
Now if $K$ be the radius of gyration in this case, we have
$\begin{array}{rl}& M{K}^2={\displaystyle \frac{5}{3}}M{R}^2\\ & orK=\sqrt{{\displaystyle \frac{5}{3}}}R\end{array}$
Now for the solid sphere it is given by
$\begin{array}{rl}& M{K}^2={\displaystyle \frac{2}{5}}M{R}^2\\ & orK=\sqrt{{\displaystyle \frac{2}{5}}}R\end{array}$
So the ratio of the radii is given by
$\begin{array}{rl}& \sqrt{{\displaystyle \frac{2}{5}}}R:\sqrt{{\displaystyle \frac{5}{3}}}R\\ & =\sqrt{6}:5\end{array}$

D) The angular momentum is the moment of linear momentum, so its SI unit would be
$kgm{s}^{-1}\times m=kg{m}^2{s}^{-1}$
From the Si unit of angular momentum, we can write its dimensions and it is given by $\left[M{L}^2{T}^{-1}\right]$.

Note:
To calculate moment of inertia, one should always remember about the axis about to be calculated. For different axes it will have different values and we need to apply parallel axes theorem and perpendicular axes theorem accordingly. Also in case radius of gyration will have the minimum value and not zero value for an axis passing through the centre of mass.