A rod of length 50 cm is pivoted at one end. It is raised such that it makes an angle of 30 degrees

Q: Why do we use the center of mass to calculate potential energy?

Because gravity acts on all parts of the rod, but mathematically, the total work done by gravity is equivalent to the work done on a point mass located at the center of mass.

Q: Why is the moment of inertia $ML^2/3$ and not $ML^2/12$?

The rod is rotating about its end (the pivot), not its center. By the parallel axis theorem, $I = ML^2/12 + M(L/2)^2 = ML^2/3$.

Q: Does the mass of the rod affect the final angular velocity?

No, the mass $M$ appears in both the potential energy and kinetic energy terms, so it cancels out completely.

Q: What if the rod was released from a vertical position?

The height change of the center of mass would be $L/2$, and you would substitute this into the energy equation to find a higher angular velocity.

Q: Why is translational kinetic energy not considered?

The rod is undergoing pure rotation about a fixed pivot. The rotational kinetic energy $\frac{1}{2}I\omega^2$ already accounts for the kinetic energy of all particles in the rod.

Question

A rod of length $50	ext{ cm}$ is pivoted at one end. It is raised such that it makes an angle of $30^\circ$ from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in $	ext{rad s}^{-1}$) will be (Take, $g = 10	ext{ ms}^{-2}$)

Options:

(A) $\frac{\sqrt{30}}{2}$

(B) $\sqrt{30}$

(C) $\frac{\sqrt{20}}{3}$

(D) $\sqrt{\frac{30}{2}}$

Sigma Percentile · Accepted Answer

The Swinging Rod: A Dance of Energy

Imagine a rigid rod pivoted at one end, held at an angle, and then released. As it swings down, it transforms its stored potential energy into a flurry of rotational motion. This classic physics problem is a beautiful demonstration of the conservation of mechanical energy.

### Visualizing the Fall

When the rod is released from an angle of $30^\circ$ above the horizontal, gravity immediately pulls it downwards. To analyze this motion, we must focus on the rod's center of mass (COM). For a uniform rod of length $L$, the COM is located exactly in the middle, at a distance of $L/2$ from the pivot.

As the rod swings from $30^\circ$ to the horizontal position ($0^\circ$), the COM drops by a certain vertical height $h$. Using basic trigonometry, we can determine this height:

$$h = \frac{L}{2} \sin 30^\circ$$

### The Energy Balance

Since there are no non-conservative forces (like friction or air resistance) doing work on the system, the total mechanical energy is conserved. This means the loss in gravitational potential energy is entirely converted into rotational kinetic energy.

The loss in potential energy is given by:

$$\Delta PE = Mgh = Mg \left( \frac{L}{2} \sin 30^\circ ight)$$

As the rod swings, it rotates about the fixed pivot. The rotational kinetic energy gained is:

$$KE = \frac{1}{2} I \omega^2$$

Here, $I$ is the moment of inertia of the rod about its end, which is $I = \frac{ML^2}{3}$.

### The Math Unfolds

Equating the loss in potential energy to the gain in kinetic energy, we get:

$$Mg \frac{L}{2} \sin 30^\circ = \frac{1}{2} \left( \frac{ML^2}{3} ight) \omega^2$$

Notice how the mass $M$ appears on both sides of the equation? It cancels out perfectly, meaning the final angular speed is independent of the rod's mass! Substituting $\sin 30^\circ = \frac{1}{2}$, the equation simplifies to:

$$g \frac{L}{4} = \frac{L^2}{6} \omega^2$$

### The Final Spin

Rearranging the equation to solve for $\omega^2$, we find:

$$\omega^2 = \frac{6g}{4L} = \frac{3g}{2L}$$

Now, we just plug in the given values: $g = 10 	ext{ m/s}^2$ and $L = 50 	ext{ cm} = 0.5 	ext{ m}$.

$$\omega^2 = \frac{3 	imes 10}{2 	imes 0.5} = \frac{30}{1} = 30$$

Taking the square root gives us the final angular speed:

$$\omega = \sqrt{30} 	ext{ rad/s}$$

And there we have it! The rod sweeps through the horizontal with an angular speed of $\sqrt{30} 	ext{ rad/s}$.

Select Answer:

Visualized Solution

$Initial Setup$

$Conservation of Energy$

$Potential Energy Loss$

$Rotational Kinetic Energy$

$Equating Energies$

$Solving for ω$

$Conclusion$

The Sigma Insight: Work and Energy in Rotational Motion

Visualizing the Fall

The Energy Balance

The Math Unfolds

The Final Spin

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conservation of Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Potential Energy Loss

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rotational Kinetic Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating Energies

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for ω

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion