Problem 592 - JEE Physics

Question

A rod of length L rotate about an axis passing through its centre and normal to its length with an angular velocity \(\omega\). If A is the cross-section and \(D\) is the density of material of rod. Find its rotational \(\mathrm{K} . \mathrm{E}\). \(\\{\mathrm{A}\\}(1 / 2) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\) \\{B \(\\}(1 / 6) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\) \(\\{C\\}(1 / 24) A L^{3} D \omega^{2}\) \(\\{\mathrm{D}\\}(1 / 12) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\)

Step-by-Step Solution

Answer

The rotational kinetic energy of the rod is \(K.E = \frac{1}{6} A L^3 D \omega^2\).

Step 1: Find the moment of inertia

To find the moment of inertia \(I\) of the rod, we can use the following formula: \[I = \int_0^L dm \, x^2\] Since the mass is uniformly distributed along the length, we can write \(dm = \frac{M}{L} dx\) , where \(M = A*L*D\) is the total mass of the rod. Now, the moment of inertia is given by: \[I = \int_0^L \frac{M}{L} x^2 dx\]

Step 2: Calculate the integral

The integral will be: \[I = \frac{M}{L} \int_0^L x^2 dx\] \[I = \frac{M}{L} \left[\frac{x^3}{3}\right]_0^L\] \[I = \frac{M}{L} \left(\frac{L^3}{3}\right)\] \[I = \frac{1}{3} M L^2\] Now, substituting the value of M, we get: \[I = \frac{1}{3} (A * L * D) L^2\] \[I = \frac{1}{3} A L^3 D\]

Step 3: Find the rotational kinetic energy

The formula for rotational kinetic energy is: \(K.E = \frac{1}{2} I \omega^2\) Now, substituting the value of I, we get: \(K.E = \frac{1}{2} (\frac{1}{3} A L^3 D) \omega^2\) \(K.E = \frac{1}{6} A L^3 D \omega^2\) So, the correct answer is \(B) (1 / 6) AL^{3} D \omega^{2}\).

Key Concepts

Moment of Inertia

The moment of inertia is a fundamental concept when discussing rotational motion. It is often compared to mass in linear motion and is sometimes called the "rotational mass". The moment of inertia determines how much torque is needed for a desired angular acceleration. It depends on how the mass of the object is distributed with respect to the axis of rotation.

In the case of a rod rotating about its center, the calculation involves integrating over the length of the rod. We consider small elements of the rod and calculate their contribution to the moment of inertia. Here's a quick breakdown of how it's done:

• For a tiny mass element, \(dm\), at a distance \(x\) from the axis of rotation, the contribution is \(dm \times x^2\).
• Since the entire mass is spread evenly along the rod, \(dm\) can be replaced by \( \frac{M}{L} dx\).
• We then integrate \(x^2\) with respect to \(x\) from 0 to \(L/2\) because we're calculating for half the length twice due to symmetry.

After the integration, putting these elements together, we get the moment of inertia of the rod about its center as: \[ I = \frac{1}{3} A L^3 D \] This moment of inertia is fundamental in calculating the rotational kinetic energy.

Angular Velocity

Angular velocity is the rate at which an object rotates around an axis. It's the rotational analog to linear velocity and provides a measure of how fast an object spins. Angular velocity \(\omega\) is typically measured in radians per second.

In this scenario, the rod spins around an axis through its center, perpendicular to its length. This means each point on the rod covers a specific angular displacement over time. This displacement and the accompanying time period characterize the angular velocity:

• Angular velocity gives insight into rotational motion dynamics and is crucial in calculating kinetic energy.
• For rotational kinetic energy \(K.E\), the formula \( \frac{1}{2} I \omega^2 \) highlights the direct role of \(\omega\) in determining the energy due to rotation.

This entire process shows the beautiful harmony between linear and rotational motion properties, enabling us to use familiar concepts like energy, velocity, and inertia in new contexts.

Density of Material

Density measures the mass of an object per unit volume and is a scalar quantity denoted by \(D\). It's crucial in deriving the mass when given the volume, like in our example where the rod's total mass is formulated as \(M = A \cdot L \cdot D\).

This property has a significant impact on rotational motion calculations because:

• Density helps specify how mass is distributed in a given volume, affecting the moment of inertia.
• For uniformly dense objects like our rod, density allows the simplification of mass calculations, directly calculating the total mass from known parameters.

Hence, it's easy to see how density intricately affects the fundamental properties of rotational motion. Whether calculating total mass, or finding how the mass varies along an object, it plays a pivotal role in the equations of motion.