Question

A rigid body is made of three identical thin rods, each with length,$\mathbf{\text{L}}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{600}\mathbf{}\mathbf{}\mathbf{\text{m}}$fastened together in the form of a letter H (Fig.$\mathbf{10}\mathbf{}\mathbf{-}\mathbf{}\mathbf{52}$). The body is free to rotate about a horizontal axis that runs along the length of one of the legs of the H. The body is allowed to fall from rest from a position in which the plane of the H is horizontal. What is the angular speed of the body when the plane of the H is vertical?

Short answer

Angular speed of the body when the plane of H is vertical is$\omega =6.06\text{rad}/\text{s}$

Step-by-step solution

Given

Length of the rod is$\text{L}=0.600\text{m}$

Explanation of concepts

We use the concept of conservation of mechanical energy. Using gravitational potential energy and rotational kinetic energy we can write the equation for angular velocity. We can find the inertia of the system, and using that inertia in the angular velocity equation we can find the answer.

1. Formulae:

$$\begin{gathered} \text{M}.\text{E}=\text{mgh}+\frac{1}{2}\text{I}{\omega }^2 \\ \text{I}={\text{MR}}^2 \end{gathered}$$

calculation

Centre of the mass of the system is at $\frac{\text{L}}{2}$and total mass of the body is M , when the system moves down the gravitational potential energy of the system decreases by $-\frac{\text{MgL}}{2}$we can write the equation of mechanical energy as,

$\text{M}.\text{E}=-\frac{\text{MgL}}{2}+\frac{1}{2}\text{I}{\omega }^2$

Initially kinetic energy of the system is zero, we get,

$$\begin{gathered} \text{M}.\text{E}=-\frac{\text{MgL}}{2}+\frac{1}{2}\text{I}{\omega }^2 \\ \frac{\text{MgL}}{2}=\frac{1}{2}\text{I}{\omega }^2 \\ {\omega }^2=\frac{\text{MgL}}{\text{I}} \\ \omega =\sqrt{\frac{\text{MgL}}{\text{I}}} \end{gathered}$$

We can find the inertia of the system about the axis passing through one of the legs of H.

Centre rod has inertial $\frac{1}{3}\left(\frac{\text{M}}{3}\right){\text{L}}^2$and the rod at the end of this rod has inertia$\left(\frac{\text{M}}{3}\right){\text{L}}^2$

We get total inertia,

$\text{I}=\frac{1}{3}\left(\frac{\text{M}}{3}\right){\text{L}}^2+\left(\frac{\text{M}}{3}\right){\text{L}}^2=\frac{{\text{ML}}^2}{9}+\frac{{\text{ML}}^2}{3}=\frac{4{\text{ML}}^2}{9}$

Plugging this value in equation of angular velocity, we get,

$$\begin{gathered} \omega =\sqrt{\frac{\text{MgL}}{\frac{4{\text{ML}}^2}{9}}} \\ \omega =\sqrt{\frac{9g}{4L}}=\sqrt{\frac{9\left(9.8\right)}{4\times 0.600}}=\sqrt{36.75} \\ \omega =6.06\raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

Final statement:

We can use the concept of conservation of mechanical energy. We can find rotational inertia of the system using conservation of energy. Using this value of inertia in the equation of mechanical energy we can find angular velocity.