Two point-like objects of masses 20 gm and 30 gm are fixed at the two ends of a rigid massless rod o

Q: Why do we need to find the center of mass?

The problem states the wire is attached to the center of mass. The wire acts as the axis of rotation for the torsional pendulum, so we must calculate the moment of inertia about this specific axis.

Q: Can I just use the total length for the moment of inertia?

No, the masses are at different distances from the axis of rotation. You must use $I = m_1r_1^2 + m_2r_2^2$, where $r_1$ and $r_2$ are their respective distances from the center of mass.

Q: Why did my answer come out wrong even though my formula was right?

The most common trap in this question is unit conversion. The masses are in grams and distances in centimeters, but the torsional constant is in SI units (N m rad⁻¹). You must convert $I$ to kg m².

Q: What is the torsional constant $C$?

It's the rotational equivalent of a spring constant. It tells you how much restoring torque the wire provides for a given angle of twist ($\tau = -C\theta$).

Q: What if the wire was attached to the center of the rod instead of the CM?

Then the axis of rotation would be at 5 cm from each mass. You would calculate the moment of inertia about that new axis, which would be higher, resulting in a lower angular frequency.

Question

Two point-like objects of masses 20 gm and 30 gm are fixed at the two ends of a rigid massless rod of length 10 cm. This system is suspended vertically from a rigid ceiling using a thin wire attached to its center of mass, as shown in the figure. The resulting torsional pendulum undergoes small oscillations. The torsional constant of the wire is $1.2 	imes 10^{-8} 	ext{ N m rad}^{-1}$. The angular frequency of the oscillations is $n 	imes 10^{-3} 	ext{ rad s}^{-1}$. The value of n is _____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a rigid rod with two unequal masses at its ends, hanging from a ceiling by a thin wire. The wire is attached exactly at the center of mass of the system. When twisted, it performs torsional oscillations.

To find the angular frequency of these oscillations, we use the standard formula for a torsional pendulum:

$$\omega = \sqrt{\frac{C}{I}}$$

We already have the torsional constant $C = 1.2 	imes 10^{-8} 	ext{ N m rad}^{-1}$, so our main task is to find the moment of inertia, $I$, about the axis of rotation.

### Locating the Center of Mass
First, let's locate the exact position of the center of mass, as this is where our axis of rotation lies. The distance of the center of mass from the heavier $30 	ext{ gm}$ mass is calculated using the center of mass formula:

$$r_1 = \frac{m_2 L}{m_1 + m_2} = \frac{20 	imes 10}{30 + 20} = 4 	ext{ cm}$$

Naturally, the distance from the $20 	ext{ gm}$ mass is the remaining length of the rod:

$$r_2 = 10 - 4 = 6 	ext{ cm}$$

### Calculating the Moment of Inertia
Now, let's calculate the moment of inertia of the system about this center of mass axis. We sum up mass times distance squared for both particles:

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

$$I = (30)(4)^2 + (20)(6)^2 = 480 + 720 = 1200 	ext{ gm cm}^2$$

**Warning:** This is where silly mistakes happen! We must convert this into standard SI units (kg m²) before plugging it into our main equation.

$$I = 1200 	imes 10^{-3} 	ext{ kg} 	imes 10^{-4} 	ext{ m}^2 = 1.2 	imes 10^{-4} 	ext{ kg m}^2$$

### The Final Calculation
We have everything we need. Let's substitute the given torsional constant $C$ and our calculated moment of inertia $I$ into the $\omega$ equation. Notice how perfectly the numbers are set up to cancel out:

$$\omega = \sqrt{\frac{1.2 	imes 10^{-8}}{1.2 	imes 10^{-4}}}$$

The $1.2$ in the numerator and denominator cancel out beautifully, leaving us with:

$$\omega = \sqrt{10^{-4}} = 10^{-2} 	ext{ rad s}^{-1}$$

The question asks for the answer in the format of $n 	imes 10^{-3}$. So, we rewrite $10^{-2}$ as $10 	imes 10^{-3}$. Comparing this, we find that the value of $n$ is exactly **10**.

Enter Numerical Value:

Visualized Solution

$m_{1}, m_{2}, and CM$

$ω = \frac{C}{I}$

$r_{1} and r_{2}$

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

$ω = \frac{1.2 \times 1 0 ^{- 8}}{1.2 \times 1 0 ^{- 4}}$

$ω = 1 0^{- 2} rad s^{- 1}$

$n = 10$

I_{new} = I_{cm} + Md^2

The Sigma Insight: Angular SHM

Analyzing the Setup Imagine a rigid rod with two unequal masses at its ends, hanging from a ceiling by a thin wire

Locating the Center of Mass First, let's locate the exact position of the center of mass, as this is where our axis of rotation lies

Calculating the Moment of Inertia Now, let's calculate the moment of inertia of the system about this center of mass axis

The Final Calculation We have everything we need

Similar Questions

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m1​,m2​, and CM

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

.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; } ω=1.2×10−41.2×10−8​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ω=10−2 rad s−1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=10