Question

A typical ten-pound car wheel has a moment of inertia of about 0.35kg *m². The wheel rotates about the axle at a constant angular speed making 70.0 full revolutions in a time interval of 5.00 seconds. What is the rotational kinetic energy K of the rotating wheel? Express answer in Joules

Answer 1

Concepts and reason

The concepts used for solving for this problem are angular velocity and the rotational kinetic energy.

First, calculate the angular velocity of the wheels by using the data provided and then use this value of angular velocity to determine the rotational kinetic energy of car.

Fundamentals

The moment of inertia is the resistance the body to accelerate and decelerate while rotating. More the moment of inertia, more the resistance provided.

The angular kinetic energy of the wheels is given by:

$\omega = \frac{{2\pi N}}{t}{\rm{ rad/s}}$ …… (1)

Here, $\omega$ is the angular velocity, N is the number of full revolutions and t is the time taken.

The rotational kinetic energy of the body is given as:

$E = \frac{1}{2}I{\omega ^2}$ …… (2)

Here, E is the rotational kinetic energy, I is the moment of inertia.

The angular velocity is determined as follows:

Substitute $70$ for N and $5{\rm{ s}}$ for t in equation (1),

$\begin{array}{c}\\\omega = \frac{{2\pi \left( {70} \right)}}{5}{\rm{ rad/s}}\\\\{\rm{ = 87}}{\rm{.96 rad/s}}\\\end{array}$

The rotational kinetic energy is determined as follows:

Substitute $87.96{\rm{ rad/s}}$ for $\omega$ and $0.35{\rm{ kg}} \cdot {{\rm{m}}^2}$ for I in equation (2),

$\begin{array}{c}\\E = \frac{1}{2}\left( {0.35} \right){\left( {87.96} \right)^2}{\rm{ J}}\\\\{\rm{ = 0}}{\rm{.5}}\left( {0.35} \right)\left( {7736.96} \right){\rm{ J}}\\\\{\rm{ = 1353}}{\rm{.96 J}}\\\end{array}$

Ans:

The rotational kinetic energy of the body is $1353.96{\rm{ J}}$ .