A thin and uniform rod of mass M and length L is held vertical on a floor with large friction. The r

Q: Why doesn't the rod slip at the bottom?

The problem states there is "large friction". This implies the static friction is sufficient to prevent any horizontal slipping, effectively acting as a fixed hinge during the fall.

Q: Why do we use the moment of inertia about the end of the rod ($ML^2/3$) instead of the center ($ML^2/12$)?

Since the rod is rotating about its fixed contact point with the floor, it is much simpler to apply the Work-Energy theorem and Torque equations directly about this fixed axis using the parallel axis theorem, which gives $I = ML^2/3$.

Q: Why does the center of mass have two components of acceleration?

The center of mass is moving in a circular path around the hinge. Any object in non-uniform circular motion has a radial acceleration (centripetal, due to changing direction) and a tangential acceleration (due to changing speed).

Q: How did we get the angles for resolving $a_r$ and $a_t$ vertically?

Geometry! The rod is at $60^\circ$ to the vertical. $a_r$ points along the rod towards the hinge, making a $60^\circ$ angle with the vertical. $a_t$ is perpendicular to the rod, so it makes a $90^\circ - 60^\circ = 30^\circ$ angle with the vertical.

Q: Why is the normal reaction so much smaller than the weight $Mg$?

The center of mass is accelerating downwards very rapidly (at $15g/16$). For an object to accelerate downwards that fast, the upward force (Normal reaction) must be very small compared to the downward force (Gravity). It's almost in free fall!

Question

A thin and uniform rod of mass $M$ and length $L$ is held vertical on a floor with large friction. The rod is released from rest so that it falls by rotating about its contact-point with the floor without slipping. Which of the following statement(s) is/are correct, when the rod makes an angle $60^\circ$ with vertical ?
[g is the acceleration due to gravity]

Options:

(A) The radial acceleration of the rod's center of mass will be $\frac{3g}{4}$

(B) The angular acceleration of the rod will be $\frac{2g}{L}$

(C) The angular speed of the rod will be $\sqrt{\frac{3g}{2L}}$

(D) The normal reaction force from the floor on the rod will be $\frac{Mg}{16}$

Sigma Percentile · Accepted Answer

### The Setup: A Falling Rod

Imagine a heavy rod standing perfectly vertical on a rough floor. Suddenly, it's released.

Because the floor has "large friction," the bottom point of the rod doesn't slip. It acts exactly like a fixed hinge.

As the rod falls, it rotates around this hinge, and its center of mass traces out a perfect circular arc. We are asked to analyze the physics of this rod at the exact moment it makes a $60^\circ$ angle with the vertical.

### Energy Conservation: Finding the Spin

To understand the forces, we first need to know how fast the rod is spinning. This is a classic application of the **Work-Energy Theorem**.

As the rod falls, it loses gravitational potential energy, which is entirely converted into rotational kinetic energy.

The center of mass of the rod is at a distance of $L/2$ from the hinge. When the rod falls from the vertical to an angle of $60^\circ$, the vertical height of the center of mass decreases.

The drop in height is $\Delta h = \frac{L}{2} - \frac{L}{2} \cos 60^\circ = \frac{L}{2}(1 - \cos 60^\circ)$.

Equating the loss in potential energy to the gain in rotational kinetic energy, we get:
$$Mg \frac{L}{2} (1 - \cos 60^\circ) = \frac{1}{2} I \omega^2$$

Using the moment of inertia of a rod about its end, $I = \frac{ML^2}{3}$, and substituting $\cos 60^\circ = \frac{1}{2}$, we can solve for the angular speed $\omega$:
$$\omega = \sqrt{\frac{3g}{2L}}$$

This confirms that **Option (C) is correct**.

Once we have the angular speed, we can immediately find the **radial acceleration** ($a_r$) of the center of mass. Since it's moving in a circle of radius $L/2$, the radial acceleration directed towards the hinge is:
$$a_r = \left(\frac{L}{2}ight) \omega^2 = \frac{3g}{4}$$

This confirms that **Option (A) is also correct**.

### The Pull of Gravity: Torque and Acceleration

Next, we need to find the angular acceleration $\alpha$. Why is the rod accelerating angularly? Because gravity is pulling it down!

Gravity acts at the center of mass, creating a torque about the hinge. The perpendicular distance from the hinge to the line of action of gravity is $\frac{L}{2} \sin 60^\circ$.

Using Newton's Second Law for rotation ($	au = I \alpha$), we write:
$$Mg \left(\frac{L}{2} \sin 60^\circight) = \left(\frac{ML^2}{3}ight) \alpha$$

Solving for $\alpha$, we get:
$$\alpha = \frac{3\sqrt{3}g}{4L}$$

This shows that Option (B) is incorrect.

Because the rod has an angular acceleration, the center of mass also has a **tangential acceleration** ($a_t$), perpendicular to the rod:
$$a_t = \left(\frac{L}{2}ight) \alpha = \frac{3\sqrt{3}g}{8}$$

### The Geometry of Motion: Resolving Accelerations

Here is where the problem gets truly beautiful—and where many students fall into a trap.

To find the normal reaction from the floor, we must analyze the forces in the vertical direction. This means we need the **net vertical acceleration** ($a_v$) of the center of mass.

The center of mass has two acceleration vectors: $a_r$ (pointing along the rod towards the hinge) and $a_t$ (pointing perpendicular to the rod downwards). Both of these vectors have downward vertical components!

By simple geometry, the radial acceleration makes a $60^\circ$ angle with the vertical, and the tangential acceleration makes a $30^\circ$ angle with the vertical.

Let's add their vertical components:
$$a_v = a_r \cos 60^\circ + a_t \cos 30^\circ$$

Substituting the values we found earlier:
$$a_v = \left(\frac{3g}{4}ight)\left(\frac{1}{2}ight) + \left(\frac{3\sqrt{3}g}{8}ight)\left(\frac{\sqrt{3}}{2}ight)$$
$$a_v = \frac{3g}{8} + \frac{9g}{16} = \frac{15g}{16}$$

The center of mass is accelerating downwards at a massive $\frac{15g}{16}$! It is almost in free fall.

### The Final Piece: Newton's Second Law

Finally, let's look at the vertical forces acting on the rod. Gravity ($Mg$) pulls down, and the Normal reaction ($N$) pushes up.

Applying Newton's Second Law in the vertical direction:
$$Mg - N = M a_v$$

Substitute our net vertical acceleration:
$$Mg - N = M \left(\frac{15g}{16}ight)$$

Solving for $N$, we get:
$$N = \frac{Mg}{16}$$

The normal reaction is just a tiny fraction of the rod's weight because the rod is accelerating downwards so rapidly. This confirms that **Option (D) is correct**.

Select Answer:

Visualized Solution

\text{Kinematics of the Falling Rod}

\text{Work-Energy Theorem}

\text{Angular Speed & Radial Acceleration}

\text{Torque Equation}

\text{Angular & Tangential Acceleration}

\text{Resolving Accelerations Vertically}

\text{Net Vertical Acceleration}

\text{Newton's Second Law (Vertical)}

\text{Normal Reaction}

The Sigma Insight: Dynamics of Rigid Body Rotation

The Setup

Energy Conservation

The Pull of Gravity

The Geometry of Motion

The Final Piece

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