18.4 Rotation of rigid body  (Page 3/4)

Problem : A particle of mass 0.1 kg is attached to one end of a light rod of length 1 m. The rod is hinged at the other end and rotates in a plane perpendicular to the axis of rotation. If the angular speed of the particle is 60 rpm, find the constant torque that can stop the rotation of the particle in one minute.

Solution : The idea here is to apply Newton's second law of rotation for a particle. For this, we need to find angular deceleration. For this, we first convert angular velocity in rad/s :

$$\begin{array}{l}{\omega }_1=60\mathrm{rpm}=60x\frac{2\pi }{60}=2\pi \mathrm{rad}/s\end{array}$$

Since torque "τ" is constant as given in the question, this is obviously the case of uniform deceleration. As such, we can apply equation of rotational motion for constant acceleration,

$$\begin{array}{l}{\omega }_1={\omega }_2+\alpha t\end{array}$$

Here,

$$\begin{array}{l}{\omega }_1=2\pi \mathrm{rad}/s;{\omega }_2=0;t=60s\end{array}$$

Putting values in the equation of motion, we have :

$$\begin{array}{l}\Rightarrow 0=2\pi +\alpha x60\\ \Rightarrow \alpha =-\frac{2\pi }{60}\mathrm{rad}/s^2\end{array}$$

Now, applying Newton's second law of motion for rotation,

$$\begin{array}{l}\tau =I\alpha \end{array}$$

Here,

$$\begin{array}{l}I=m{r}^2=0.1x{1}^2=0.1\mathrm{kg}-m^2\end{array}$$

Dropping the sign of deceleration and putting values in the expression of torque, we have the magnitude of torque :

$$\begin{array}{l}\tau =I\alpha =0.1x\frac{2\pi }{60}=0.01N-m\end{array}$$

Problem : Three particles each of mass "m" are situated at the vertices of an equilateral triangle OAB of length "a" as shown in the figure. Calculate moment of inertia (i) about an axis passing through "O" and perpendicular to the plane of triangle (ii) about axis Ox and (iii) about axis Oy.

Moment of inertia

Solution : We shall make use of the formulae of moment of inertia for discrete particles in each of the cases :

(i) about an axis passing through "O" and perpendicular to the plane of triangle

Here, distances of three particles from the axis are :

$$\begin{array}{l}{r}_O=0;r_A=a;r_B=a\end{array}$$

The moment of inertia about an axis through "O" and perpendicular to the plane of triangle is :

$$\begin{array}{l}I=\sum m_i{r_i}^2=m_o{r_o}^2+m_A{r_A}^2+m_B{r_B}^2\end{array}$$

$$\begin{array}{l}\Rightarrow I=mx0+m{a}^2+m{a}^2=2m{a}^2\end{array}$$

(ii) about axis Ox

Here distances of three particles from the axis of rotation are :

$$\begin{array}{l}{r}_O=0;r_A=\frac{\surd 3a}{2};r_B=0\end{array}$$

$$\begin{array}{l}I=\sum m_i{r_i}^2=m_o{r_o}^2+m_A{r_A}^2+m_B{r_B}^2\end{array}$$

$$\begin{array}{l}\Rightarrow I=mx0+m{\left(\frac{\surd 3a}{2}\right)}^2+mx0=\frac{3m{a}^2}{4}\end{array}$$

(iii) about axis Oy

Here distances of three particles from the axes are :

$$\begin{array}{l}{r}_O=0,r_A=\frac{a}{2},r_B=a\end{array}$$

$$\begin{array}{l}I=\sum m_i{r_i}^2=m_o{r_o}^2+m_A{r_A}^2+m_B{r_B}^2\end{array}$$

$$\begin{array}{l}\Rightarrow I=mx0+\frac{m{a}^2}{4}+mx{a}^2=\frac{5m{a}^2}{4}\end{array}$$

This example illustrates how choice of axis changes moment of inertia i.e. the inertia of the system of particles to rotation. This is expected as change in the reference axis actually changes distribution of mass about axis of rotation.