Chapter 8, Problem 78GP

(a)

To determine

To Find: The angular acceleration of the rod at the moment of release.

Explanation of Solution

.

Introduction:

The angular acceleration is defined as the rate of change of velocity and it is designated as $\alpha$ . It is measured in radians per second squared ( $\text{rad}/{\text{s}}^{\text{2}}$ ). Angular acceleration depends on the change in angular velocity $\omega$ .

The torque in the rod can be determined by the following equation,

  $\tau =I\alpha$

The moment of inertia of the rod is given by,

  $\begin{array}{l}I=\frac{M{L}^2}{3}\\ \tau =Mg\frac{L}{2}\end{array}$

Here, M is the mass of the rod and L is the length of the rod.

Equating both torque equations

  $\begin{array}{l}\tau =Mg\frac{L}{2}=\frac{M{L}^2}{3}\alpha \\ \alpha =\frac{3g}{2L}\end{array}$

Conclusion:

At the moment of release, the angular accelerations are $\frac{3g}{2L}$ .

(b)

To determine

To Find: The linear acceleration of the tip of the rod.

Explanation of Solution

.

Introduction:

The angular acceleration and the linear acceleration are directly proportional to each other. The greater the linear (tangential) acceleration is, the larger the angular acceleration is, and vice versa.

The angular acceleration is given by the equation,

  $I\alpha \text{=}\left(\frac{wl}{2}\right)$

Where

  $\alpha \text{=}\left(\frac{m{l}^2}{3}\right)$

  $w=mg$

On solving, the angular acceleration of the rod is given as

  $\alpha \text{=}\left(\frac{3g}{2l}\right)$

The linear acceleration of the tip of the rod is given as

  $\begin{array}{l}a=l\alpha \\ a=l\left(\frac{3g}{2l}\right)\\ a=\left(\frac{3g}{2}\right)\end{array}$

Conclusion:

The linear acceleration of the tip of the rod is $\left(\frac{3g}{2}\right)$ .