A typical ten-pound car wheel has a moment of inertia of about 0.35 kg*m^2. The wheel rotates about the axle at a constant angular speed making 45.0 full revolutions in a time interval of 7.00 sec .
What is the rotational kinetic energy of the rotating wheel?
To calculate the rotational kinetic energy of the rotating wheel, we can use the formula:
Rotational Kinetic Energy = (1/2) * Moment of Inertia * (Angular Speed)^2
Given:
Moment of Inertia (I) = 0.35 kg*m^2
Angular Speed (ω) = (45.0 revolutions) / (7.00 seconds)
First, let's convert the angular speed from revolutions per second to radians per second. Since one revolution is equal to 2π radians, we can do the following conversion:
Angular Speed (ω) = (45.0 revolutions / 7.00 seconds) * (2π radians/1 revolution)
Now, we can calculate the rotational kinetic energy:
Rotational Kinetic Energy = (1/2) * 0.35 kg*m^2 * (Angular Speed)^2
Substituting the values, we get:
Rotational Kinetic Energy = (1/2) * 0.35 kg*m^2 * [(45.0 revolutions / 7.00 seconds) * (2π radians/1 revolution)]^2
Simplifying the equation, we have:
Rotational Kinetic Energy ≈ (1/2) * 0.35 kg*m^2 * (9.4248 radians/second)^2
Rotational Kinetic Energy ≈ (1/2) * 0.35 kg*m^2 * 88.826 radians^2/second^2
Rotational Kinetic Energy ≈ 15.530 kg*m^2*radians^2/second^2
Therefore, the rotational kinetic energy of the rotating wheel is approximately 15.530 joules.
See
http://www.jiskha.com/display.cgi?id=
1256540146