Question

The end B of a uniform rod AB of length $l$ which makes an angle $\theta$ with the floor is pulled with a velocity $V_0$ as shown. At the instant when $\theta=37^0$

• A. Velocity of end A is $\frac{4}{3}V_0$
• B. Angular velocity of the rod is $\frac{5V_0}{3l}$
• C. Velocity of C.M of rod is $\frac{5V_0}{6}$
• D. Kinetic energy of the rod is $\frac{25}{54}mV_0^2$

Answer 1

Answer: (A), (B), (C), (D)

A, B, C, D

Answer 2

The problem involves a uniform rod sliding between a vertical wall and a horizontal floor. We are given the velocity of end B and asked to find various kinematic quantities and the kinetic energy of the rod at a specific instant.

Let's set up a coordinate system with the origin O at the corner where the floor and wall meet. Let the end A of the rod be at $(0, y_A)$ and the end B be at $(x_B, 0)$. The length of the rod is $l$. From the diagram, the angle $\theta$ is between the rod and the floor (x-axis). Thus, we can write the coordinates of A and B as: $x_B = l \cos\theta$ $y_A = l \sin\theta$

We are given that end B is pulled with a velocity $V_0$ along the positive x-axis. So, $\frac{dx_B}{dt} = V_0$. We need to find the values at $\theta = 37^\circ$. We use the approximations: $\sin 37^\circ \approx 3/5$ and $\cos 37^\circ \approx 4/5$.

1. Velocity of end A ($V_A$)

Differentiate the equation for $y_A$ with respect to time: $\frac{dy_A}{dt} = l \cos\theta \frac{d\theta}{dt}$ Let $V_A = \frac{dy_A}{dt}$ and $\omega = \frac{d\theta}{dt}$ (angular velocity). So, $V_A = l \cos\theta \cdot \omega$. Differentiate the equation for $x_B$ with respect to time: $\frac{dx_B}{dt} = l (-\sin\theta) \frac{d\theta}{dt}$ $V_0 = -l \sin\theta \cdot \omega$ From this, we can find $\omega$: $\omega = -\frac{V_0}{l \sin\theta}$ Now substitute $\omega$ into the expression for $V_A$: $V_A = l \cos\theta \left(-\frac{V_0}{l \sin\theta}\right) = -V_0 \frac{\cos\theta}{\sin\theta} = -V_0 \cot\theta$ The negative sign indicates that end A is moving downwards. The magnitude of velocity of end A is $|V_A| = V_0 \cot\theta$. At $\theta = 37^\circ$: $\cot 37^\circ = \frac{\cos 37^\circ}{\sin 37^\circ} = \frac{4/5}{3/5} = \frac{4}{3}$ $|V_A| = V_0 \left(\frac{4}{3}\right) = \frac{4}{3}V_0$. So, option A is correct.

2. Angular velocity of the rod ($\omega$)

From the derivation above: $|\omega| = \frac{V_0}{l \sin\theta}$ At $\theta = 37^\circ$: $\sin 37^\circ = 3/5$ $|\omega| = \frac{V_0}{l (3/5)} = \frac{5V_0}{3l}$. So, option B is correct. (The direction of rotation is clockwise, as $\theta$ is decreasing).

3. Velocity of C.M of the rod ($V_{CM}$)

The center of mass (C.M) of a uniform rod is at its midpoint. Let its coordinates be $(x_C, y_C)$. $x_C = \frac{x_B}{2} = \frac{l \cos\theta}{2}$ $y_C = \frac{y_A}{2} = \frac{l \sin\theta}{2}$ The velocity components of the C.M are: $V_{Cx} = \frac{dx_C}{dt} = \frac{1}{2} \frac{dx_B}{dt} = \frac{V_0}{2}$ $V_{Cy} = \frac{dy_C}{dt} = \frac{1}{2} \frac{dy_A}{dt} = \frac{V_A}{2} = \frac{1}{2} \left(-V_0 \cot\theta\right) = -\frac{V_0}{2} \cot\theta$ The magnitude of the velocity of the C.M is: $V_{CM} = \sqrt{V_{Cx}^2 + V_{Cy}^2} = \sqrt{\left(\frac{V_0}{2}\right)^2 + \left(-\frac{V_0}{2} \cot\theta\right)^2}$ $V_{CM} = \frac{V_0}{2} \sqrt{1 + \cot^2\theta} = \frac{V_0}{2} \sqrt{\csc^2\theta} = \frac{V_0}{2} |\csc\theta|$ At $\theta = 37^\circ$: $\csc 37^\circ = \frac{1}{\sin 37^\circ} = \frac{1}{3/5} = \frac{5}{3}$ $V_{CM} = \frac{V_0}{2} \left(\frac{5}{3}\right) = \frac{5V_0}{6}$. So, option C is correct.

4. Kinetic energy of the rod (K.E)

The kinetic energy of a rigid body is the sum of its translational kinetic energy (due to the C.M) and its rotational kinetic energy (about the C.M): $K.E = \frac{1}{2} m V_{CM}^2 + \frac{1}{2} I_{CM} \omega^2$ For a uniform rod of mass $m$ and length $l$, the moment of inertia about its center of mass (perpendicular to the rod) is $I_{CM} = \frac{1}{12} m l^2$. Substitute the values we found for $V_{CM}$ and $\omega$: $V_{CM} = \frac{5V_0}{6}$ $\omega = \frac{5V_0}{3l}$ $K.E = \frac{1}{2} m \left(\frac{5V_0}{6}\right)^2 + \frac{1}{2} \left(\frac{1}{12} m l^2\right) \left(\frac{5V_0}{3l}\right)^2$ $K.E = \frac{1}{2} m \frac{25V_0^2}{36} + \frac{1}{24} m l^2 \frac{25V_0^2}{9l^2}$ $K.E = \frac{25mV_0^2}{72} + \frac{25mV_0^2}{216}$ To sum these fractions, find a common denominator, which is 216: $K.E = \frac{3 \times 25mV_0^2}{3 \times 72} + \frac{25mV_0^2}{216}$ $K.E = \frac{75mV_0^2}{216} + \frac{25mV_0^2}{216}$ $K.E = \frac{75mV_0^2 + 25mV_0^2}{216} = \frac{100mV_0^2}{216}$ Simplify the fraction by dividing the numerator and denominator by 4: $K.E = \frac{25mV_0^2}{54}$. So, option D is correct.

All options A, B, C, and D are correct.