Question

A rod is kept and lowest pt is moving with speed v find velocity as function oftime

Answer 1

Velocity of center of mass: $\vec{V}_C(t) = \left(\frac{v}{2}, -\frac{(x_0 + vt)v}{2\sqrt{L^2 - (x_0 + vt)^2}}\right)$, Angular velocity: $\omega(t) = -\frac{v}{\sqrt{L^2 - (x_0 + vt)^2}}$

Answer 2

This problem describes the motion of a rod sliding against a vertical wall and a horizontal floor. We need to find the velocity of the rod as a function of time, given that its lowest point moves with a constant speed $v$. For a rigid body, its motion is fully described by the velocity of its center of mass and its angular velocity.

1. Define the Coordinate System and Variables: Let the length of the rod be $L$.
Let the lowest point of the rod be $A$, which is on the horizontal floor. Its coordinates are $(x, 0)$.
Let the highest point of the rod be $B$, which is on the vertical wall. Its coordinates are $(0, y)$.

2. Establish the Constraint Equation: Since the rod has a fixed length $L$, the distance between its ends $A$ and $B$ must remain constant. Using the Pythagorean theorem:
$x^2 + y^2 = L^2$

3. Relate Velocities using the Constraint: We are given that the lowest point moves with a speed $v$. This means $\frac{dx}{dt} = v$.
To find the velocity of the top point $\frac{dy}{dt}$, we differentiate the constraint equation with respect to time $t$:
$\frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(L^2)$
$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$
Substitute $\frac{dx}{dt} = v$:
$2xv + 2y \frac{dy}{dt} = 0$
$xv + y \frac{dy}{dt} = 0$
$\frac{dy}{dt} = -\frac{xv}{y}$
The negative sign indicates that as $x$ increases (lowest point moves away from the wall), $y$ decreases (highest point slides down the wall).

4. Express Positions as Functions of Time: Let $x_0$ be the initial position of the lowest point at $t=0$. Since $\frac{dx}{dt} = v$ is constant:
$x(t) = x_0 + vt$
Now, substitute $x(t)$ into the constraint equation to find $y(t)$:
$y(t) = \sqrt{L^2 - x(t)^2} = \sqrt{L^2 - (x_0 + vt)^2}$

5. Velocity of the Center of Mass ($V_C$): For a uniform rod, the center of mass $C$ is at the midpoint of the rod. Its coordinates are $(x_C, y_C) = \left(\frac{x}{2}, \frac{y}{2}\right)$.
The velocity of the center of mass is $\vec{V}_C = \left(\frac{dx_C}{dt}, \frac{dy_C}{dt}\right)$.
$\frac{dx_C}{dt} = \frac{1}{2} \frac{dx}{dt} = \frac{v}{2}$
$\frac{dy_C}{dt} = \frac{1}{2} \frac{dy}{dt} = -\frac{xv}{2y}$
Substituting $x(t)$ and $y(t)$:
$\vec{V}_C(t) = \left(\frac{v}{2}, -\frac{(x_0 + vt)v}{2\sqrt{L^2 - (x_0 + vt)^2}}\right)$

6. Angular Velocity ($\omega$): Let $\theta$ be the angle the rod makes with the horizontal floor.
Then $x = L \cos \theta$ and $y = L \sin \theta$.
Differentiate $x = L \cos \theta$ with respect to time:
$\frac{dx}{dt} = -L \sin \theta \frac{d\theta}{dt}$
We know $\frac{dx}{dt} = v$ and $\frac{d\theta}{dt} = \omega$ (angular velocity).
So, $v = -L \sin \theta \omega$
$\omega = -\frac{v}{L \sin \theta}$
Since $\sin \theta = \frac{y}{L}$, substitute this into the expression for $\omega$:
$\omega = -\frac{v}{L \left(\frac{y}{L}\right)} = -\frac{v}{y}$
Now, substitute $y(t)$:
$\omega(t) = -\frac{v}{\sqrt{L^2 - (x_0 + vt)^2}}$
The negative sign indicates that the angle $\theta$ is decreasing as the rod slides down.

Conclusion: The velocity of the rod, as a function of time, can be described by its center of mass velocity and its angular velocity.

• Velocity of the Center of Mass:
  $\vec{V}_C(t) = \left(\frac{v}{2}, -\frac{(x_0 + vt)v}{2\sqrt{L^2 - (x_0 + vt)^2}}\right)$

• Angular Velocity:
  $\omega(t) = -\frac{v}{\sqrt{L^2 - (x_0 + vt)^2}}$

Here, $L$ is the length of the rod, $v$ is the constant speed of the lowest point, and $x_0$ is the initial horizontal position of the lowest point at $t=0$.