Question

A long, uniform rod of length L and mass M is pivoted about a frictionless, horizontal pin through one end. The rod is nudged from rest in a vertical position as shown in Figure P10.73. At the instant the rod is horizontal, find (a) its angular speed,

(b) the magnitude of its angular acceleration,

(c) the x and y components of the acceleration of its center of mass, and

(d) the components of the reaction force at the pivot.

Short answer

(a). Its angular speed is$\omega =\sqrt{\frac{3g}{L}}$.

(b) The magnitude of its angular acceleration is$\alpha =\frac{3}{2}\frac{g}{L}$.

(c) The x and y components of the acceleration of its center of mass is$a_x=-\frac{3g}{2},a_y=-\frac{3g}{4}$.

(d) The components of the reaction force at the pivot are,$R_x=-\frac{3mg}{2}, R_y=\frac{mg}{4}$.

Step-by-step solution

Angular speed

Angular speed or rotational speed, also known as angular frequency vector, is a vector measurement of the rate of rotation that refers to how fast an object rotates or rotates relative to another point, i.e. over time, how fast the angular position or orientation of an object changes.

 Step 2: Finding the angular speed

(a)

The equation of moment of inertia for the uniform rod is$I=\frac{1}{3}m{L}^2$.

The energy conservation equation is $mg\frac{L}{2}=\frac{1}{2}I{\omega }^2$.

Substitute I in the above equation,

$mg\frac{L}{2}=\frac{1}{2} \left(\frac{1}{3}m{L}^2\right)\times {\omega }^2$

Rewriting the above for$\omega$,

$\begin{array}{c}\omega =\sqrt{\frac{mgL}{\frac{1}{3}m{L}^2}}\\ \omega =\sqrt{\frac{3g}{L}}\end{array}$

Its angular speed is$\omega =\sqrt{\frac{3g}{L}}$.

Finding the angular acceleration

(b)

The torque formula is,

$\begin{array}{c}\tau =I·\alpha \\ I·\alpha =\frac{L}{2}mg\end{array}$

Rewriting the formula for$\alpha$,

$\alpha =\frac{L·m·g}{2·I}$

Where,$I=\frac{1}{3}m{L}^2$.

$\begin{array}{rcl}\alpha & =& \frac{L·m·g}{2·\frac{1}{3}m{L}^2}\\ \alpha & =& \frac{3}{2}\frac{g}{L}\end{array}$

The magnitude of its angular acceleration is$\alpha =\frac{3}{2}\frac{g}{L}$.

Finding the acceleration for x and y components

(c)

The formula for the horizontal component of acceleration,

$\begin{array}{c}{a}_x=-\frac{L}{2}{\omega }^2\\ =-\frac{L}{2}\frac{3g}{L}\end{array}$

The horizontal component acceleration (x) is $a_x=-\frac{3g}{2}$.

The formula for the vertical component of acceleration,

$\begin{array}{c}{a}_y=-\frac{L}{2}\alpha \\ =-\frac{L}{2}\frac{3}{2}\frac{g}{L}\end{array}$

The vertical component of acceleration (y) is$a_y=-\frac{3g}{4}$.

 Find components of the reaction force at the pivot:

(d)

$\begin{array}{rcl}{R}_x& =& m·a_x\\ R_x& =& -\frac{3mg}{2}\end{array}$

and,

$\begin{array}{rcl}{R}_y& =& mg-\frac{3}{4}mg\\ R_y& =& \frac{mg}{4}\end{array}$