Question 500: Centre of Mass and Rotational Motion

Question

Two circular loops P and Q of radii ' r ' and ' nr ' are made respectively from a uniform wire. Moment of inertia of loop Q about its axis is 4 times that of loop $P$ about its axis. The value of $n$ is

Accepted Answer

Let the mass of wire used to make loop $P$ be $m_P$ and for loop $Q$ be $m_Q$.
Since both loops are made from uniform wire of the
same length
, let's first find the relationship between their masses and radii.
Step 1: Equate lengths of wires
Length of wire in $P = 2\pi r$
Length of wire in $Q = 2\pi (nr)$
Since the wires are of equal length:
$$ 2\pi r_P = 2\pi r_Q \implies r_P = r_Q $$
But here, the loops are of radii $r$ and $nr$, so the length of the wire for $P$ is $2\pi r$, and for $Q$ it is $2\pi n r$.
Let mass per unit length = $\lambda$.
So,
$$ m_P = \lambda \cdot 2\pi r $$
$$ m_Q = \lambda \cdot 2\pi n r $$
Step 2: Write the moments of inertia
The moment of inertia of a thin ring (loop) about its axis:
$$ I = m r^2 $$
So,
$$ I_P = m_P r^2 = (\lambda \cdot 2\pi r) \cdot r^2 = 2\pi\lambda r^3 $$
$$ I_Q = m_Q (nr)^2 = (\lambda \cdot 2\pi n r) \cdot (n^2 r^2) = 2\pi\lambda n r \cdot n^2 r^2 = 2\pi\lambda n^3 r^3 $$
Step 3: Use the condition $I_Q = 4I_P$
From above,
$$ I_Q = 2\pi\lambda n^3 r^3 $$
$$ I_P = 2\pi\lambda r^3 $$
Given,
$$ I_Q = 4 I_P $$
So,
$$ 2\pi\lambda n^3 r^3 = 4 \cdot 2\pi\lambda r^3 $$
$$ n^3 = 4 $$
$$ n = 4^{1/3} = (2^2)^{1/3} = 2^{2/3} $$
Final Answer:
$$ \boxed{(2)^{2/3}} $$
So, the correct option is
Option B
.

Answer

$(2)^{-2 / 3}$

Answer

$(2)^{2 / 3}$

Answer

$\sqrt{2}$

Answer

$2^{1 / 3}$

Centre of Mass and Rotational Motion

Two circular loops P and Q of radii ' r ' and ' nr ' are made respectively from a uniform wire. Moment of inertia of loop Q about its axis is 4 times that of loop P about its axis. The value of n is

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