Question

Particles of masses 1 g, 2 g, 3 g.......... 100 g are kept at the marks 1 cm, 2 cm, 3 cm,......, 100 cm respectively on a metre scale. Find the moment of inertia of the system of particles about a perpendicular bisector of the metre scale.

• A. 2.5 kg-m²
• B. 1 kg-m²
• C. 0.43 kg-m²
• D. 2 kg-m²

Answer 1

Answer: (C)

0.43 kg-m²

Answer 2

The problem asks us to find the moment of inertia of a system of particles about a perpendicular bisector of a metre scale.

1. Define the system and axis of rotation:

• A metre scale is 100 cm long.
• The particles are placed at marks 1 cm, 2 cm, ..., 100 cm.
• The mass of the particle at mark $i$ cm is $m_i = i$ g.
• The axis of rotation is the perpendicular bisector of the metre scale. This means the axis passes through the 50 cm mark.

2. Set up coordinates and convert units to SI:

Let's place the origin of our coordinate system at the 50 cm mark.

• The position of the $i$-th particle is $x_i = i$ cm.
• The mass of the $i$-th particle is $m_i = i$ g $= i \times 10^{-3}$ kg.
• The distance of the $i$-th particle from the axis (at 50 cm) is $r_i = |x_i - 50|$ cm.

Since we need $r_i^2$, we can use $r_i^2 = (x_i - 50)^2$ cm$^2$. $r_i = (i - 50)$ cm $= (i - 50) \times 10^{-2}$ m. So, $r_i^2 = (i - 50)^2 \times (10^{-2})^2$ m$^2 = (i - 50)^2 \times 10^{-4}$ m$^2$.

3. Formula for Moment of Inertia:

For a system of discrete particles, the moment of inertia $I$ about an axis is given by the sum of $m_k r_k^2$ for all particles: $I = \sum_{i=1}^{100} m_i r_i^2$

4. Substitute and set up the summation:

$I = \sum_{i=1}^{100} (i \times 10^{-3} \text{ kg}) \times ((i - 50)^2 \times 10^{-4} \text{ m}^2)$ $I = 10^{-7} \sum_{i=1}^{100} i (i - 50)^2$ kg-m$^2$

Expand the term inside the summation: $i(i - 50)^2 = i(i^2 - 100i + 2500) = i^3 - 100i^2 + 2500i$

So, $I = 10^{-7} \left( \sum_{i=1}^{100} i^3 - 100 \sum_{i=1}^{100} i^2 + 2500 \sum_{i=1}^{100} i \right)$

5. Use standard summation formulas:

We need the sum of the first $N$ integers, squares of integers, and cubes of integers, where $N=100$:

• $\sum_{i=1}^{N} i = \frac{N(N+1)}{2}$
• $\sum_{i=1}^{N} i^2 = \frac{N(N+1)(2N+1)}{6}$
• $\sum_{i=1}^{N} i^3 = \left( \frac{N(N+1)}{2} \right)^2$

Calculate each sum for $N=100$:

• $\sum_{i=1}^{100} i = \frac{100(100+1)}{2} = \frac{100 \times 101}{2} = 50 \times 101 = 5050$
• $\sum_{i=1}^{100} i^2 = \frac{100(100+1)(2 \times 100+1)}{6} = \frac{100 \times 101 \times 201}{6} = 50 \times 101 \times 67 = 338350$
• $\sum_{i=1}^{100} i^3 = (5050)^2 = 25502500$

6. Substitute the sums back into the moment of inertia equation:

$I = 10^{-7} \left( 25502500 - 100 \times 338350 + 2500 \times 5050 \right)$ $I = 10^{-7} \left( 25502500 - 33835000 + 12625000 \right)$ $I = 10^{-7} \left( (25502500 + 12625000) - 33835000 \right)$ $I = 10^{-7} \left( 38127500 - 33835000 \right)$ $I = 10^{-7} \left( 4292500 \right)$ $I = 0.42925$ kg-m$^2$

Rounding to two decimal places, $I \approx 0.43$ kg-m$^2$.

The final answer is $\boxed{\text{0.43 kg-m²}}$.

Explanation of the solution:

The moment of inertia is calculated as the sum of $m r^2$ for all particles. The masses are $i$ grams and positions are $i$ cm. The axis is at 50 cm. Convert all units to SI (kg and m). The distance from the axis for the $i$-th particle is $(i-50)$ cm. The total moment of inertia becomes a sum involving $i(i-50)^2$. This sum is expanded into terms of $i^3$, $i^2$, and $i$. Standard summation formulas for the sum of integers, squares, and cubes are used for $N=100$. The calculated sum is then multiplied by the unit conversion factor ($10^{-7}$) to get the result in kg-m$^2$.

Answer:

The final answer is 0.43 kg-m². The correct option is (C).