Question

A cricket mat is rolled loosely in the form of a cylinder of radius 2 m. The mass of the mat is 50 kg. The mat is then rolled tightly so that the radius of the cylinder formed reduces to 3/4th of its original value. Find the ratio of the moment of inertia of the mat in the two cases.
A) $1:3$
B) $4:3$
C) $3:5$
D) $16:9$

Answer 1

The tightly or loosely rolled mat makes up a solid cylinder. The moment of inertia of a cylinder is proportional to the square of its radius.

Formulas used:
-The moment of inertia of a cylinder is given by, $I = \dfrac{{M{R^2}}}{2}$ where $M$ is the mass of the cylinder and $R$ is the radius of the cylinder.

Complete step by step answer.
Step 1: List the data provided in the question.
It is given that the cricket mat of mass $M = 50{\text{kg}}$ is rolled into a cylinder.
Let’s now consider the two cases - when the mat was rolled loosely and rolled tightly.
Case 1:
The mat is rolled loosely into a cylinder of a radius ${R_1} = 2{\text{m}}$ .
Case 2:
The mat is rolled tightly into a cylinder whose radius is 3/4th of its original value.
Let ${R_2}$ be the radius of the cylinder in the second case, then we have ${R_2} = \dfrac{3}{4}{R_1}$
Step 2: Express the moment of inertia of the mat in the first case.
We know that the moment of inertia of a cylinder of mass $M$ and radius $R$ is given by,
$I = \dfrac{{M{R^2}}}{2}$ ----------(1)
Since the mat is rolled in the form of a cylinder we can use equation (1) to express the initial moment of inertia of the mat.
Let ${I_1}$ be the moment of inertia of the mat when it was loosely rolled.

Then the moment of inertia of the mat in the first case is ${I_1} = \dfrac{{MR_1^2}}{2}$ ------- (2)
Step 3: Express the moment of inertia of the mat in the second case.
In the second case, the radius of the cylinder is given by, ${R_2} = \dfrac{3}{4}{R_1}$
We can use equation (1) to express the moment of inertia of the mat when it was rolled tightly.
Let ${I_2}$ be the moment of inertia of the mat in this case.
Then we have, ${I_2} = \dfrac{{MR_2^2}}{2}$
Substituting for ${R_2} = \dfrac{3}{4}{R_1}$ in the above expression we get, ${I_2} = \dfrac{{M{{\left( {\dfrac{{3{R_1}}}{4}} \right)}^2}}}{2}$
On simplifying we get, the moment of inertia of the mat in the second case as
${I_2} = \dfrac{{9MR_1^2}}{{32}}$ ---------- (3)
Step 4: Using equations (2) and (3) obtain the required ratio.
Equation (2) gives ${I_1} = \dfrac{{MR_1^2}}{2}$ and equation (3) gives ${I_2} = \dfrac{{9MR_1^2}}{{32}}$
Substitute values for $M = 50{\text{kg}}$ and ${R_1} = 2{\text{m}}$ in the equations (2) and (3).
Then ${I_1} = \dfrac{{MR_1^2}}{2} = \dfrac{{50 \times 4}}{2}$ and ${I_2} = \dfrac{{9MR_1^2}}{{32}} = \dfrac{{9 \times 50 \times 4}}{{32}}$
Taking the ratio ${I_1}:{I_2}$ , we get $\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{\dfrac{{50 \times 4}}{2}}}{{\dfrac{{9 \times 50 \times 4}}{{32}}}}$
Cancel out the similar terms in the above equation to get, $\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{16}}{9}$

Therefore, the ratio of the moment of inertia of the mat in the two cases is $16:9$ .

Note: The ratio can also be found out without substituting the values.
Equations (2) and (3) represent expressions for ${I_1}$ and ${I_2}$ respectively.
Then taking the ratio ${I_1}:{I_2}$ , we get $\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{\dfrac{{MR_1^2}}{2}}}{{\dfrac{{9MR_1^2}}{{32}}}}$
Now simply cancel out the similar terms to obtain the required ratio.
We then have, $\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{\dfrac{1}{2}}}{{\dfrac{9}{{32}}}}$ and on simplifying we get, ${I_1}:{I_2} = 16:9$