Question

A solid cylinder of mass 20kg rotates about its axis with angular speed of $100rad{s^{ - 1}}$. The radius of the cylinder is 0.25m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum about its axis?

Answer 1

The kinetic energy associated with the rotation of the cylinder is given by the formula –
$K = \dfrac{1}{2}I{\omega ^2}$
where I – moment of inertia about the axis $\omega$- angular velocity
Angular momentum is defined as the product of moment of inertia and the angular velocity. Angular momentum, L –
$L = I\omega$

Complete step-by-step answer:
Step 1: Calculate moment of inertia.
The moment of inertia is a quantity that expresses the tendency of resisting the rotation of the body about its axis. The moment of inertia for a solid cylinder of mass m and radius r rotating about its axis, is given by –
$I = \dfrac{{m{r^2}}}{2}$
Substituting the values of m and r in the formula, we can calculate the moment of inertia.
$I = \dfrac{{m{r^2}}}{2} \\\ m = 20kg \\\ r = 0.25m \\\ I = \dfrac{{20 \times {{\left( {0.25} \right)}^2}}}{2} \\\ Solving, \\\ I = \dfrac{{{{20}}10 \times 0.0625}}{{{2}}} \\\ I = 0.625kg - {m^2} \\\$
Step 2: Use the value of moment of inertia to calculate the kinetic energy

$$K = \dfrac{1}{2}I{\omega ^2} \\\ \omega = 100rad{s^{ - 1}} \\\ Substituting, \\\ K = \dfrac{1}{2} \times 0.625 \times {100^2} \\\ Solving, \\\ K = \dfrac{1}{2} \times 0.625 \times 10000 \\\ K = \dfrac{{6250}}{2} = 3125J \\\$$

Step 3: Calculate the angular momentum
Angular momentum, $L = I\omega$
Substituting,
$L = 0.625 \times 100 \\\ Solving, \\\ L = 62.5kg{m^2}/\sec \\\$
Hence, the answer is Kinetic energy, $K = 3125J$ and Angular momentum, $L = 62.5kg{m^2}/\sec$

Note: There is also an alternate method to find the angular momentum using the equation for kinetic energy.
The rotational kinetic energy is given by –
$K = \dfrac{1}{2}I{\omega ^2}$
We know that, $L = I\omega$
So, by substituting the value of L in the above equation for kinetic energy, we get –
$K = \dfrac{1}{2}\left( {I\omega } \right)\omega$
Substituting,
$K = \dfrac{1}{2}L\omega \\\ rearranging, \\\ L = \dfrac{{2K}}{\omega } \\\$
This shortcut can be quickly used in your competitive exams if the inputs given are kinetic energy and angular velocity only. So, you would save time to calculate the moment of inertia because you have to remember how to solve for the moment of inertia in particular cases. This shortcut can quickly help you obtain the value for angular momentum.