Question

A uniform rod AB of length $l$ can slide between a horizontal surface and a slope of inclination $\phi$ as shown in the figure. If the end A is moving with uniform velocity $v_A$ away from the wall, find the:

(a) velocity of the end B. (b) angular velocity of the rod. (c) velocity of the mass center.

Answer 1

a) Velocity of end B: $\frac{v_A \cos \theta}{\sin(\theta + \phi)}$ (down the incline)

b) Angular velocity of the rod: $\frac{v_A \cos \phi}{l \sin(\theta + \phi)}$ (clockwise)

c) Velocity of the mass center: $\frac{v_A \cos \phi}{2 \sin(\theta + \phi)}$ (along the rod)

Answer 2

The problem describes a uniform rod AB of length $l$ sliding between a vertical wall and an inclined slope. The end A is on the vertical wall and moves upwards with a uniform velocity $v_A$. The end B is on the inclined slope, which makes an angle $\phi$ with the horizontal. Let $\theta$ be the angle the rod makes with the vertical wall.

1. Setup and Kinematic Relations:

Let the origin (0,0) be the corner where the vertical wall meets the horizontal ground. The coordinates of end A are $(x_A, y_A)$. Since A is on the vertical wall, $x_A = 0$. The velocity of A is $\vec{v}_A = v_A \hat{j}$ (upwards along the y-axis, as shown in the figure). So, $\frac{dx_A}{dt} = 0$ and $\frac{dy_A}{dt} = v_A$.

Let the coordinates of end B be $(x_B, y_B)$. The length of the rod AB is $l$. From the geometry (refer to the figure): $x_B = l \sin \theta$ $y_A - y_B = l \cos \theta \implies y_B = y_A - l \cos \theta$

Differentiating these position equations with respect to time to find velocities: $\frac{dx_B}{dt} = v_{Bx} = l \cos \theta \frac{d\theta}{dt}$ $\frac{dy_B}{dt} = v_{By} = \frac{dy_A}{dt} - l (-\sin \theta) \frac{d\theta}{dt} = v_A + l \sin \theta \frac{d\theta}{dt}$

Let the angular velocity of the rod be $\omega = \frac{d\theta}{dt}$. So, $v_{Bx} = l \omega \cos \theta$ $v_{By} = v_A + l \omega \sin \theta$

2. Constraint on Velocity of End B:

End B slides along the inclined plane. The inclined plane makes an angle $\phi$ with the horizontal. The velocity of B must be directed along the incline. From the figure, as A moves up, B moves down the incline. The slope of the incline is $-\tan \phi$ (if we consider $y$ decreasing as $x$ increases). Therefore, the components of $\vec{v}_B$ must satisfy: $v_{By} = -v_{Bx} \tan \phi$

Substitute the expressions for $v_{Bx}$ and $v_{By}$: $v_A + l \omega \sin \theta = -(l \omega \cos \theta) \tan \phi$ $v_A = -l \omega \sin \theta - l \omega \cos \theta \tan \phi$ $v_A = -l \omega (\sin \theta + \cos \theta \tan \phi)$ $v_A = -l \omega \left(\sin \theta + \frac{\cos \theta \sin \phi}{\cos \phi}\right)$ $v_A = -l \omega \left(\frac{\sin \theta \cos \phi + \cos \theta \sin \phi}{\cos \phi}\right)$ $v_A = -l \omega \frac{\sin(\theta + \phi)}{\cos \phi}$

Part (b) Angular velocity of the rod:

From the above equation, the magnitude of the angular velocity $\omega$ is: $\omega = \left| \frac{v_A \cos \phi}{-l \sin(\theta + \phi)} \right| = \frac{v_A \cos \phi}{l \sin(\theta + \phi)}$ The negative sign indicates that $\theta$ is decreasing as A moves up, which is consistent with the geometry. So, the direction of angular velocity is clockwise.

Part (a) Velocity of the end B:

We have $v_{Bx} = l \omega \cos \theta$ and $v_{By} = -v_{Bx} \tan \phi$. Substitute the expression for $\omega$: $v_{Bx} = l \left( \frac{v_A \cos \phi}{l \sin(\theta + \phi)} \right) \cos \theta = \frac{v_A \cos \phi \cos \theta}{\sin(\theta + \phi)}$ $v_{By} = -\left( \frac{v_A \cos \phi \cos \theta}{\sin(\theta + \phi)} \right) \tan \phi = -\frac{v_A \cos \phi \cos \theta \sin \phi}{\sin(\theta + \phi) \cos \phi} = -\frac{v_A \sin \phi \cos \theta}{\sin(\theta + \phi)}$

The velocity of end B is $\vec{v}_B = v_{Bx} \hat{i} + v_{By} \hat{j}$. The magnitude of $\vec{v}_B$ is: $v_B = \sqrt{v_{Bx}^2 + v_{By}^2} = \sqrt{\left(\frac{v_A \cos \phi \cos \theta}{\sin(\theta + \phi)}\right)^2 + \left(-\frac{v_A \sin \phi \cos \theta}{\sin(\theta + \phi)}\right)^2}$ $v_B = \frac{v_A \cos \theta}{\sin(\theta + \phi)} \sqrt{\cos^2 \phi + \sin^2 \phi} = \frac{v_A \cos \theta}{\sin(\theta + \phi)}$ The direction of $\vec{v}_B$ is down the incline, making an angle $\phi$ below the horizontal.

Part (c) Velocity of the mass center:

For a uniform rod, the mass center C is at the midpoint of AB. Coordinates of C: $(x_C, y_C)$. $x_C = \frac{x_A + x_B}{2} = \frac{0 + l \sin \theta}{2} = \frac{l}{2} \sin \theta$ $y_C = \frac{y_A + y_B}{2}$

Velocity of mass center $\vec{v}_C = v_{Cx} \hat{i} + v_{Cy} \hat{j}$. $v_{Cx} = \frac{dx_C}{dt} = \frac{l}{2} \cos \theta \frac{d\theta}{dt} = \frac{l}{2} \omega \cos \theta$ $v_{Cy} = \frac{dy_C}{dt} = \frac{1}{2} \left( \frac{dy_A}{dt} + \frac{dy_B}{dt} \right) = \frac{1}{2} (v_A + v_{By})$

Substitute $\omega$, $v_{Bx}$, and $v_{By}$: $v_{Cx} = \frac{l}{2} \left( \frac{v_A \cos \phi}{l \sin(\theta + \phi)} \right) \cos \theta = \frac{v_A \cos \phi \cos \theta}{2 \sin(\theta + \phi)}$ $v_{Cy} = \frac{1}{2} \left( v_A - \frac{v_A \sin \phi \cos \theta}{\sin(\theta + \phi)} \right) = \frac{v_A}{2} \left( \frac{\sin(\theta + \phi) - \sin \phi \cos \theta}{\sin(\theta + \phi)} \right)$ $v_{Cy} = \frac{v_A}{2} \left( \frac{\sin \theta \cos \phi + \cos \theta \sin \phi - \sin \phi \cos \theta}{\sin(\theta + \phi)} \right) = \frac{v_A \sin \theta \cos \phi}{2 \sin(\theta + \phi)}$

The velocity of the mass center is $\vec{v}_C = \frac{v_A \cos \phi \cos \theta}{2 \sin(\theta + \phi)} \hat{i} + \frac{v_A \sin \theta \cos \phi}{2 \sin(\theta + \phi)} \hat{j}$. The magnitude of $\vec{v}_C$ is: $v_C = \sqrt{v_{Cx}^2 + v_{Cy}^2} = \sqrt{\left(\frac{v_A \cos \phi \cos \theta}{2 \sin(\theta + \phi)}\right)^2 + \left(\frac{v_A \sin \theta \cos \phi}{2 \sin(\theta + \phi)}\right)^2}$ $v_C = \frac{v_A \cos \phi}{2 \sin(\theta + \phi)} \sqrt{\cos^2 \theta + \sin^2 \theta} = \frac{v_A \cos \phi}{2 \sin(\theta + \phi)}$ The direction of $\vec{v}_C$ is given by the angle $\alpha_C$ with the horizontal: $\tan \alpha_C = \frac{v_{Cy}}{v_{Cx}} = \frac{\frac{v_A \sin \theta \cos \phi}{2 \sin(\theta + \phi)}}{\frac{v_A \cos \phi \cos \theta}{2 \sin(\theta + \phi)}} = \frac{\sin \theta}{\cos \theta} = \tan \theta$. So, $\alpha_C = \theta$. This means the velocity of the mass center is directed along the rod (from B towards A).

Subject: Physics Chapter: Rotational Motion Topic: Kinematics of Rigid Bodies / Relative Velocity in Rigid Bodies