Question 16: Centre of Mass and Rotational Motion

Question

Two rings of radius $$R$$ and $$n R$$ made of same material have the ratio of moment of inertia about an axis passing through centre is $$1: 64$$. The value of $$n$$ is

Accepted Answer

To determine the value of $$n$$, let's analyze the problem by considering the moment of inertia for each ring about an axis passing through its center. For a ring, the moment of inertia about an axis through its center can be expressed as:
$$I = mR^2$$
Where:
$$I$$ is the moment of inertia
$$m$$ is the mass of the ring
$$R$$ is the radius of the ring
Given that both rings are made of the same material, they will have the same density. Let's denote the moment of inertia of the first ring (radius $$R$$) as $$I_1$$ and the moment of inertia of the second ring (radius $$nR$$) as $$I_2$$. According to the problem:
$$\frac{I_1}{I_2} = \frac{1}{64}$$
For the first ring (radius $$R$$):
$$I_1 = m_1 R^2$$
For the second ring (radius $$nR$$):
$$I_2 = m_2 (nR)^2 = m_2 n^2 R^2$$
Assuming both rings are made of the same material and have the same mass per unit length, the mass of the second ring with radius $$nR$$ will be $$n$$ times the mass of the first ring (since circumference scales linearly with radius):
$$m_2 = n m_1$$
Then we can write:
$$I_2 = n m_1 (nR)^2 = n m_1 n^2 R^2 = n^3 m_1 R^2$$
Using the ratio of the moments of inertia:
$$\frac{I_1}{I_2} = \frac{m_1 R^2}{n^3 m_1 R^2} = \frac{1}{n^3} = \frac{1}{64}$$
The above equality simplifies to:
$$n^3 = 64$$
$$n = \sqrt[3]{64} = 4$$
Thus, the value of $$n$$ is:
Option C

Answer

2

Answer

2$$\sqrt2$$

Answer

4

Answer

1/2

Centre of Mass and Rotational Motion

Two rings of radius $R and n R made of same material have the ratio of moment of inertia about an axis passing through centre is 1: 64. The value of n$ is

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