Question

Two uniform solid cylinders, each rotating about its central (longitudinal) axis at 235rad/s, have the same mass of 1.25kg but differ in radius. What is the rotational kinetic energy of (a) the smaller cylinder, of radius 0.25m, and (b) the larger cylinder, of radius 0.75m?

Short answer

1. Rotational kinetic energy of smaller cylinder is, $K_s=1.08kJ$
2. Rotational kinetic energy of larger cylinder is $K_l=9.71kJ$

Step-by-step solution

Understanding the given information

1. Mass of small cylinder, m_s = 1.25kg
2. Mass of large cylinder, m_l = 1.25kg
3. Angular velocity of small cylinder, ${\omega }_s=235rad/s$
4. Angular velocity of small cylinder, ${\omega }_l=235rad/s$
5. Radius of small cylinder, $R_s=0.25m$
6. Radius of large cylinder,$R_l=0.25m$

Concept and Formula used in the given question

We can find the moment of inertia of each cylinder using the given data. Using this value of rotational inertia in a rotational kinetic energy formula, we will find rotational kinetic energy of smaller and larger cylinder by using the formula given below.

1. Rotational Kinetic energy,$\mathit{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{l}{\mathit{\omega }}^{\mathbf{2}}$
2. Moment of inertia of solid cylinder,$\mathit{l}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{R}}^{\mathbf{2}}$

(a) Calculation for the rotational kinetic energy of the small cylinder

To calculate rotational kinetic energy of the small cylinder, we need to find its moment of inertia.

Moment of inertia of small solid cylinder

$\begin{array}{rcl}{l}_s& =& \frac{1}{2}{m}_s{R}_s^2\\ & =& -\frac{1}{2}\times 1.25kg\times {\left(0.25m^2\right)}^2\\ & =& 0.039kg.m^2\end{array}$

Using this in rotational kinetic energy formula

$\begin{array}{rcl}{K}_s& =& \frac{1}{2}{l}_s{\omega }_s^2\\ & =& \frac{1}{2}\times 0.039kg/m^2\times {\left(235rad/s\right)}^2\\ & =& 1076J\\ & =& 1.08kJ\end{array}$

Rotational kinetic energy of small cylinder is 1.08kJ.

(b) Calculation for the rotational kinetic energy of the large cylinder

To calculate rotational kinetic energy of the large cylinder, we need to find its moment of inertia.

Moment of inertia of large solid cylinder

$\begin{array}{rcl}{l}_l& =& \frac{1}{2}{m}_l{R}_l^2\\ & =& \frac{1}{2}\times 1.25kg\times {\left(0.75m^2\right)}^2\\ & =& 0.3515kg.m^2\end{array}$

Using this in rotational kinetic energy formula

$\begin{array}{rcl}{K}_l& =& \frac{1}{2}{l}_l{\omega }_l^2\\ & =& \frac{1}{2}\times 0.3515kg.m^2\times {\left(235rad/s\right)}^2\\ & =& 9705.79J\\ & =& 9.71kJ\end{array}$

Rotational kinetic energy of large cylinder is 9.71kJ.