Question:
A particle executes uniform circular motion with an angular momentum $L$. If its kinetic energy is doubled and the angular frequency is halved, then its angular momentum becomes:
A particle executes uniform circular motion with an angular momentum $L$. If its kinetic energy is doubled and the angular frequency is halved, then its angular momentum becomes:
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Using the relationship $L = \frac{2K}{\omega}$, we can see how the variables scale:
$$L \propto \frac{K}{\omega}$$
If $K$ is multiplied by 2 and $\omega$ is divided by 2, the scaling factor for $L$ is:
$$\text{Factor} = \frac{2}{1/2} = 4$$
This allows you to find the answer without full algebraic substitution.
Updated On: Jun 10, 2026
- $2\text{ L}$
- $4\text{ L}$
- $L/2$
- $L/4$
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The Correct Option is B
Solution and Explanation
Concept:
For a particle rotating in a circular trajectory, its rotational kinetic energy ($K$) and angular momentum ($L$) can be written using its moment of inertia ($I$) and angular frequency/velocity ($\omega$):
• Rotational Kinetic Energy: $K = \frac{1}{2}I\omega^2$
• Angular Momentum: $L = I\omega$
We can find a direct relationship between kinetic energy, angular momentum, and angular velocity by substituting $I = \frac{L}{\omega}$ into the kinetic energy formula: $$K = \frac{1}{2} \left(\frac{L}{\omega}\right) \omega^2 = \frac{1}{2}L\omega$$ Rearranging this formula allows us to express the angular momentum $L$ directly as: $$L = \frac{2K}{\omega}$$
Step 1: Let the initial state parameters be defined as follows: $$\text{Initial Kinetic Energy} = K_1 = K$$ $$\text{Initial Angular Frequency} = \omega_1 = \omega$$ $$\text{Initial Angular Momentum} = L_1 = L = \frac{2K}{\omega}$$
Step 2: According to the problem statement, the system values are modified such that: $$\text{New Kinetic Energy } K_2 = 2K$$ $$\text{New Angular Frequency } \omega_2 = \frac{\omega}{2}$$
Step 3: Substitute these modified parameters into the formula to find the new angular momentum $L_2$: $$L_2 = \frac{2K_2}{\omega_2} = \frac{2(2K)}{\left(\frac{\omega}{2}\right)}$$ Simplifying the complex fraction by moving the denominator's factor of 2 to the numerator: $$L_2 = \frac{4K}{\frac{\omega}{2}} = 4K \cdot \frac{2}{\omega} = 8 \cdot \left(\frac{K}{\omega}\right)$$ We can rewrite this expression to compare it directly with our original angular momentum formula $L = \frac{2K}{\omega}$: $$L_2 = 4 \cdot \left( \frac{2K}{\omega} \right) = 4 \cdot L_1 = 4L$$ Therefore, the final angular momentum becomes $4\text{ L}$, matching Option (B).
• Rotational Kinetic Energy: $K = \frac{1}{2}I\omega^2$
• Angular Momentum: $L = I\omega$
We can find a direct relationship between kinetic energy, angular momentum, and angular velocity by substituting $I = \frac{L}{\omega}$ into the kinetic energy formula: $$K = \frac{1}{2} \left(\frac{L}{\omega}\right) \omega^2 = \frac{1}{2}L\omega$$ Rearranging this formula allows us to express the angular momentum $L$ directly as: $$L = \frac{2K}{\omega}$$
Step 1: Let the initial state parameters be defined as follows: $$\text{Initial Kinetic Energy} = K_1 = K$$ $$\text{Initial Angular Frequency} = \omega_1 = \omega$$ $$\text{Initial Angular Momentum} = L_1 = L = \frac{2K}{\omega}$$
Step 2: According to the problem statement, the system values are modified such that: $$\text{New Kinetic Energy } K_2 = 2K$$ $$\text{New Angular Frequency } \omega_2 = \frac{\omega}{2}$$
Step 3: Substitute these modified parameters into the formula to find the new angular momentum $L_2$: $$L_2 = \frac{2K_2}{\omega_2} = \frac{2(2K)}{\left(\frac{\omega}{2}\right)}$$ Simplifying the complex fraction by moving the denominator's factor of 2 to the numerator: $$L_2 = \frac{4K}{\frac{\omega}{2}} = 4K \cdot \frac{2}{\omega} = 8 \cdot \left(\frac{K}{\omega}\right)$$ We can rewrite this expression to compare it directly with our original angular momentum formula $L = \frac{2K}{\omega}$: $$L_2 = 4 \cdot \left( \frac{2K}{\omega} \right) = 4 \cdot L_1 = 4L$$ Therefore, the final angular momentum becomes $4\text{ L}$, matching Option (B).
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