Question 586: Centre of Mass and Rotational Motion

Question

Two rotating bodies $$P$$ and $$Q$$ of masses '$$\mathrm{m}$$' and '$$2 \mathrm{~m}$$' with moment of inertia $$I_P$$ and $$I_Q\left(I_Q > I_P\right)$$ have equal Kinetic energy of rotation. If $$\mathrm{L}_P$$ and $$\mathrm{L}_Q$$ be their angular momenta respectively then

Accepted Answer

Here's a step-by-step explanation to determine the relationship between the angular momenta of the two rotating bodies:
Understanding the Concepts
Rotational Kinetic Energy:
The kinetic energy of a rotating body is given by:
$$K.E. = \frac{1}{2} I \omega^2$$
where:
$$I$$ is the moment of inertia
$$\omega$$ is the angular velocity
Angular Momentum:
The angular momentum of a rotating body is given by:
$$L = I \omega$$
Analyzing the Problem
Equal Kinetic Energy:
The problem states that the kinetic energies of the two bodies are equal:
$$ \frac{1}{2} I_P \omega_P^2 = \frac{1}{2} I_Q \omega_Q^2$$
Moment of Inertia Relationship:
We are given that $$I_Q > I_P$$.
Solving for Angular Velocities:
From the equal kinetic energy equation, we can see that:
$$\omega_Q^2 = \frac{I_P}{I_Q} \omega_P^2$$
Since $$I_Q > I_P$$, this implies $$\omega_Q^2  I_P$$ and $$\omega_Q  L_P$$.
Conclusion
Therefore, the correct answer is
Option D:  $$L_Q > L_P$$
.

Answer

$$\mathrm{L}_{\mathrm{Q}}=0$$

Answer

$$\mathrm{L}_Q=\mathrm{L}_P$$

Answer

$$\mathrm{L}_Q<\mathrm{L}_{\mathrm{P}}$$

Answer

$$\mathrm{L}_Q > \mathrm{L}_{\mathrm{P}}$$

Centre of Mass and Rotational Motion

Two rotating bodies $P and Q of masses '\mathrm{m}' and '2 \mathrm{~m}' with moment of inertia I_P and I_Q\left(I_Q > I_P\right) have equal Kinetic energy of rotation. If \mathrm{L}_P and \mathrm{L}_Q$ be their angular momenta respectively then

Choose the correct answer: