Question

A rod of mass 𝑀, length 𝐿 and non-uniform mass per unit length \(λ(x)=\frac{3Mx^2}{L^3},\) is held horizontally by a pivot, as shown in the figure, and is free to move in the plane of the figure. For this rod, which of the following statements are true?

• A. Moment of inertia of the rod about an axis passing through the pivot is \(\frac{3}{5} \) 𝑀𝐿²
• B. Moment of inertia of the rod about an axis passing through the pivot is \(\frac{1}{3}\) ML²
• C. Torque on the rod about the pivot i \(\frac{3}{4}\) MgL
• D. If the rod is released, the point at a distance\(\frac{2L}{3}\) from the pivot will fall with acceleration g

Answer 1

The Correct Option is A, C

To solve the problem, let's analyze each statement step by step:

1. The mass per unit length of the rod is given as \(λ(x)=\frac{3Mx^2}{L^3}\). To find the mass of a small element \(dx\) at a distance \(x\) from the pivot, we have:

\(dm = λ(x) \cdot dx = \frac{3Mx^2}{L^3} \cdot dx\)

2. The moment of inertia \(dI\) of this small element about the pivot is given by:

\(dI = x^2 \cdot dm = x^2 \cdot \frac{3Mx^2}{L^3} \cdot dx\)

3. Integrate \(dI\) from \(0\) to \(L\) to get the total moment of inertia \(I\):

\(I = \int_{0}^{L} x^2 \cdot \frac{3Mx^2}{L^3} \, dx = \frac{3M}{L^3} \cdot \int_{0}^{L} x^4 \, dx\)

\( = \frac{3M}{L^3} \cdot \left[\frac{x^5}{5}\right]_{0}^{L} = \frac{3M}{L^3} \cdot \frac{L^5}{5} = \frac{3}{5}ML^2\)

• Therefore, the moment of inertia of the rod about an axis passing through the pivot is \(\frac{3}{5}ML^2\). This statement is correct.

4. To find the torque \(\tau\) on the rod about the pivot due to gravity, first find the force on an element:

\(df = dm \cdot g = \frac{3Mgx^2}{L^3} \cdot dx\)

5. The torque due to this elemental force is:

\(d\tau = x \cdot df = x \cdot \frac{3Mgx^2}{L^3} \cdot dx = \frac{3Mgx^3}{L^3} \cdot dx\)

6. Integrate from \(0\) to \(L\) to get total torque:

\(\tau = \int_{0}^{L} \frac{3Mgx^3}{L^3} \, dx = \frac{3Mg}{L^3} \cdot \int_{0}^{L} x^3 \, dx\)

\( = \frac{3Mg}{L^3} \cdot \left[\frac{x^4}{4}\right]_{0}^{L} = \frac{3Mg}{L^3} \cdot \frac{L^4}{4} = \frac{3}{4}MgL\)

• Hence, the torque on the rod about the pivot is \(\frac{3}{4}MgL\). This statement is correct.

7. Regarding the statement about the point at a distance \(\frac{2L}{3}\) falling with acceleration g, that would be true if it were in free fall. However, due to the rotational nature and pivot constraints, this is not accurate; the acceleration will be different.

Thus, the two correct statements are:

• Moment of inertia of the rod about an axis passing through the pivot is \(\frac{3}{5}ML^2\).
• Torque on the rod about the pivot is \(\frac{3}{4}MgL\).