Question

Two point masses of \(0.3 \mathrm{~kg}\) and \(0.7 \mathrm{~kg}\) are fixed at the ends of a rod of length \(1.4 \mathrm{~m}\) and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of ..... \(\\{\mathrm{A}\\} 0.4 \mathrm{~m}\) from mass of \(0.3 \mathrm{~kg}\) \\{B \(0.98 \mathrm{~m}\) from mass of \(0.3 \mathrm{~kg}\) \\{C\\} \(0.7 \mathrm{~m}\) from mass of \(0.7 \mathrm{~kg}\) \\{D \(\\} 0.98 \mathrm{~m}\) from mass of \(0.7 \mathrm{~kg}\)

Short answer

The correct answer is \(\\{\mathrm{B}\\} 0.98 \mathrm{~m}\) from mass of \(0.3 \mathrm{~kg}\).

Step-by-step solution

Determine the moment of inertia

Moment of inertia, denoted by I, is the property of an object to resist change in its angular motion. For a point mass rotating about an axis parallel to its length, the moment of inertia is given by the mass times the square of the distance from the axis to the point mass, i.e. \(I = md^2\). In our case, we have two point masses, so the net moment of inertia would be the sum of the individual moments of inertia. For mass \(m_1 = 0.3 \mathrm{~kg}\), let the distance from the axis to the mass be d. The moment of inertia for this mass is \(I_1 = m_1 d^2\). For mass \(m_2 = 0.7 \mathrm{~kg}\), since the total length of the rod is \(1.4 \mathrm{~m}\), the distance from the axis to this mass would be \(1.4 \mathrm{~-} d\). The moment of inertia for this mass is \(I_2 = m_2 (1.4 - d)^2\). The net moment of inertia is given by \(I = I_1 + I_2\).

Differentiate the moment of inertia with respect to distance

Next, we need to minimize the moment of inertia I with respect to the distance d. To do this, we'll differentiate I with respect to d and set the derivative equal to 0. Therefore, \(\frac{dI}{dd} = \frac{d}{dd}(0.3d^2 + 0.7(1.4-d)^2) = 0\).

Solve for distance

Now, we need to solve the equation obtained in step 2 for the distance d. \(\frac{d}{dd}(0.3d^2 + 0.7(1.4-d)^2) = 0.6d - 1.4(0.7) + 1.4(0.7)d = 0\) Simplifying and solving for d, we get: \(d(d(0.6 + 0.7\cdot1.4) = 1.4(0.7)\) \(d = \frac{1.4(0.7)}{0.6 + 0.7(1.4)}\) \(d = 0.98 \mathrm{~m}\) According to the answer, the distance of 0.98 m is from mass \(0.3 \mathrm{~kg}\), which corresponds to option B. Thus, the correct answer is: \(\\{\mathrm{B}\\} 0.98 \mathrm{~m}\) from mass of \(0.3 \mathrm{~kg}\)