Question

Let the moment of inertia of a hollow cylinder of length 30 cm (inner radius 10 cm and outer radius 20 cm), about its axis be I. The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also I, is:
(A) 12 cm (B) 16 cm
(C) 14 cm (D) 18 cm

Answer 1

Given :

Length of hollow cylinder :

=30 cm

Inner radius of hollow cylinder :

=10 cm

Outer radius of hollow cylinder ;

= 20 cm

Mass of both hollow and thin cylinders are same .

To Find :

Radius of thin cylinder = ?

Solution :

For hollow cylinder :

$dm = \rho dv$

$dm =\frac{M}{\pi h (20^{2}-10^{2}) } \times 2\pi r. dr .h$

$dm= \frac{M}{30 \times 10} \times 2r .dr$

And the moment of inertia I for this hollow cylinder will be given as :

$dI=dm\times r^{2}$

Now integrating both sides ;

$\int dI= \frac{2M}{300} \times \int r^{3} dr$

Herr the limit of r is from 10 cm to 20 cm , so on further it can be written as :

$I= \frac{2M}{300} \times \frac{r^{4} }{4}$

Now applying the limits it comes as :

$I= \frac{2M}{300} \times \frac{(20^{4}-10^{4}) }{4}$

$I= \frac{M(20^{2}+10^{2}) }{2}$  (1)

Now the formula for the moment of inertia for thin cylinder is :

$I= MR^{2}$   - (2)

So on equating the above equations i.e.  eq (1) and (2) we get :

$MR^{2} = \frac{M}{2} \times( 20^{2} +10^{2})$

$R^{2} =\frac{(20^{2}+ 10^{2}) }{2}$

$R^{2} = \frac{500}{2}$

$R= \sqrt{250}$

R ≈ 16 cm

So the radius of thin cylinder is 16 cm .