Two masses m and m/2 are connected at the two ends of a massless rigid rod of length l. The rod is s

Q: Why is the wire attached at the center of mass?

Attaching the wire at the center of mass ensures that the system undergoes pure rotational simple harmonic motion without any translational swinging (like a pendulum). It keeps the axis of rotation stationary.

Q: Is the tension the same throughout the rod?

Yes! For a massless rigid rod, the tension must be uniform to provide the exact required centripetal force for both masses. If you calculate the centripetal force for the $m/2$ mass, you will find it is exactly equal to the force required for the mass $m$.

Q: Why do we use maximum angular velocity to find the tension?

The question specifically asks for the tension when the rod passes through its mean position. In simple harmonic motion, the kinetic energy and velocity (both linear and angular) are at their absolute maximum at the mean position.

Q: What is the difference between $\omega_{\text{SHM}}$ and $\omega_{\text{max}}$?

$\omega_{\text{SHM}}$ is the constant angular frequency of the oscillation itself (how fast it completes a cycle). $\omega_{\text{max}}$ is the physical, instantaneous angular speed of the rod as it rotates, which peaks at the mean position.

Q: Does gravity affect the tension in this specific setup?

No. The rod is rotating in a horizontal plane. Gravity acts vertically downwards, which is perpendicular to the tension acting horizontally along the length of the rod. The wire supports the weight, while the rod's tension only handles the centripetal force.

Question

Two masses $m$ and $m/2$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system (see figure). Because of torsional constant $k$, the restoring torque is $	au = k	heta$ for angular displacement $	heta$. If the rod is rotated by $	heta_0$ and released, the tension in it when it passes through its mean position will be

Options:

(A) $\frac{2k	heta_0^2}{l}$

(B) $\frac{k	heta_0^2}{l}$

(C) $\frac{3k	heta_0^2}{l}$

(D) $\frac{k	heta_0^2}{2l}$

Sigma Percentile · Accepted Answer

The problem of a torsional pendulum with a rod-mass system is a classic test of your understanding of rotational mechanics and simple harmonic motion. It beautifully weaves together the concepts of center of mass, moment of inertia, angular kinematics, and centripetal force. Let's break down this elegant problem step by step.

### Analyzing the Setup: The Center of Mass

Before we can analyze any rotation, we must find the axis of rotation. The problem states that the rod is suspended at its center of mass. Let's determine exactly where this point lies.

We have a rod of length $l$ with mass $m$ at one end and mass $m/2$ at the other. Let's place the mass $m$ at the origin ($x = 0$) and the mass $m/2$ at $x = l$. The position of the center of mass, $x_{	ext{cm}}$, is given by:

$$x_{	ext{cm}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$$

Substituting our values:

$$x_{	ext{cm}} = \frac{m(0) + (m/2)(l)}{m + m/2} = \frac{ml/2}{3m/2} = \frac{l}{3}$$

This tells us that the center of mass is located at a distance of $l/3$ from the mass $m$. Consequently, the distance from the mass $m/2$ to the center of mass is $l - l/3 = 2l/3$. The wire is attached exactly at this point, and the system will rotate about this vertical axis.

### The Master Equation: Moment of Inertia

To understand the torsional oscillation, we need the system's resistance to twisting, which is its moment of inertia ($I$) about the axis of rotation (the center of mass). The rod is massless, so we only consider the two point masses.

$$I = m_1 r_1^2 + m_2 r_2^2$$

Plugging in the distances we just found:

$$I = m\left(\frac{l}{3}ight)^2 + \frac{m}{2}\left(\frac{2l}{3}ight)^2$$

Now, let's carefully expand and simplify this expression:

$$I = m\left(\frac{l^2}{9}ight) + \frac{m}{2}\left(\frac{4l^2}{9}ight)$$

$$I = \frac{ml^2}{9} + \frac{2ml^2}{9} = \frac{3ml^2}{9} = \frac{ml^2}{3}$$

This is the total moment of inertia of our rod-mass system.

### Angular Kinematics: The Maximum Velocity

When the rod is twisted by an initial angle $	heta_0$ and released, the restoring torque $	au = k	heta$ causes it to execute angular simple harmonic motion (SHM). The angular frequency $\omega_{	ext{SHM}}$ of this motion is determined by the torsional constant $k$ and the moment of inertia $I$:

$$\omega_{	ext{SHM}} = \sqrt{\frac{k}{I}}$$

In SHM, the velocity is maximum when the system passes through its mean (equilibrium) position. The maximum angular velocity $\omega_{	ext{max}}$ is the product of the angular amplitude $	heta_0$ and the angular frequency:

$$\omega_{	ext{max}} = 	heta_0 \omega_{	ext{SHM}} = 	heta_0 \sqrt{\frac{k}{I}}$$

At this instant, both masses are moving in circular arcs at their maximum linear speeds. For mass $m$, which is at a radius $r = l/3$, the maximum linear speed $v_{	ext{max}}$ is:

$$v_{	ext{max}} = r \omega_{	ext{max}} = \frac{l}{3} 	heta_0 \sqrt{\frac{k}{I}}$$

### Final Calculation: The Invisible Thread of Tension

Why is there tension in the rod? As the masses swing through the mean position, they are moving in a circle. Any object moving in a circle requires a centripetal force directed towards the center of rotation. In this setup, the rigid rod provides this necessary centripetal force through its internal tension $T$.

Let's calculate the required centripetal force for the mass $m$:

$$F_c = \frac{m v_{	ext{max}}^2}{r} = m r \omega_{	ext{max}}^2$$

This centripetal force is exactly equal to the tension $T$ in the rod at that point. Substituting our expressions for $r$ and $\omega_{	ext{max}}$:

$$T = m \left(\frac{l}{3}ight) \left(	heta_0 \sqrt{\frac{k}{I}}ight)^2$$

$$T = \frac{m l 	heta_0^2 k}{3 I}$$

Now, for the grand finale, we substitute the moment of inertia $I = ml^2/3$ that we calculated earlier into our tension equation:

$$T = \frac{m l 	heta_0^2 k}{3 \left(\frac{ml^2}{3}ight)}$$

Notice the beautiful algebraic cancellation! The $3$ in the numerator and denominator cancel out. The mass $m$ cancels out. One factor of $l$ cancels out. We are left with a remarkably simple and elegant final expression:

**$$T = \frac{k 	heta_0^2}{l}$$**

This is the tension in the rod as it whips through its mean position. The correct option is (b).

Select Answer:

Visualized Solution

$x_{cm}$

$r_{1} and r_{2}$

$I = m_{1} r_{1}^{2} + m_{2} r_{2}^{2}$

$I_{setup}$

$I_{final}$

$ω_{max}$

$v_{max}$

$F_{c} = T$

$T_{setup}$

$T_{final}$

The Sigma Insight: Angular SHM

Analyzing the Setup

The Master Equation

Angular Kinematics

Final Calculation

Similar Questions

Select Answer:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r1​ and r2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I=m1​r12​+m2​r22​

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

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

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

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

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

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

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