Question

Two discs rotating about their respective axis of rotation with angular speeds $2 \text{ rads}^{-1}$ and $5 \text{ rads}^{-1}$ are brought into contact such that their axes of rotation coincide. Now, the angular speed of the system becomes $4 \text{ rads}^{-1}$. If the moment of inertia of the second disc is $1 \times 10^{-3} \text{ kg m}^2$, then the moment of inertia of the first disc (in $\text{kg m}^2$) is}

• A. $0.25 \times 10^{-3}$
• B. $1.5 \times 10^{-3}$
• C. $1.25 \times 10^{-3}$
• D. $0.75 \times 10^{-3}$
• E. $0.5 \times 10^{-3}$

Hint:

Think of this as the rotational version of a perfectly inelastic collision ($m_1v_1 + m_2v_2 = (m_1+m_2)v_f$). The math is identical, just using rotational variables!

Answer 1

Answer: (E)

Concept: This problem is based on the Principle of Conservation of Angular Momentum.
• Conservation of Angular Momentum: If no external torque acts on a system, the total angular momentum remains constant ($L_i = L_f$).
• Angular Momentum ($L$): Defined as $L = I\omega$, where $I$ is the moment of inertia and $\omega$ is the angular speed.
• Final System: When two discs are joined, they rotate together with a common angular speed $\omega_f$, and the total moment of inertia is the sum of their individual moments ($I_1 + I_2$).

Step 1: Set up the conservation equation. Initial total angular momentum ($L_i$) is the sum of the individual momenta: \[ L_i = I_1\omega_1 + I_2\omega_2 \] Final total angular momentum ($L_f$) after they are brought together: \[ L_f = (I_1 + I_2)\omega_f \] According to the principle: \[ I_1\omega_1 + I_2\omega_2 = (I_1 + I_2)\omega_f \]

Step 2: Substitute values and solve for $I_1$. Given: $\omega_1 = 2 \text{ rads}^{-1}$, $\omega_2 = 5 \text{ rads}^{-1}$, $\omega_f = 4 \text{ rads}^{-1}$, and $I_2 = 1 \times 10^{-3} \text{ kg m}^2$. \[ I_1(2) + (1 \times 10^{-3})(5) = (I_1 + 1 \times 10^{-3})(4) \] \[ 2I_1 + 5 \times 10^{-3} = 4I_1 + 4 \times 10^{-3} \] Rearranging terms: \[ 5 \times 10^{-3} - 4 \times 10^{-3} = 4I_1 - 2I_1 \] \[ 1 \times 10^{-3} = 2I_1 \] \[ I_1 = \frac{1}{2} \times 10^{-3} = 0.5 \times 10^{-3} \text{ kg m}^2 \]