when 100 J of work is done on a stationary flywheel ( that is otherwise free to rotate in place), its angular speed increases from 60 rev/min. What is its moment of inertia?
When 100j of work is done upon a fly wheel,its angular speed increases from 60rev/min to 180rev/min.whar is the moment of inertia
To find the moment of inertia of the flywheel, we can use the equation:
ΔKE = 1/2 * I * Δω^2
Where:
ΔKE is the change in kinetic energy
I is the moment of inertia of the flywheel
Δω is the change in angular speed
In this case, we know that 100 J of work is done on the flywheel, and the angular speed increases from 60 rev/min. We need to convert the angular speed from rev/min to rad/s before we can use the equation.
1 revolution = 2π radians
1 minute = 60 seconds
So, to convert from rev/min to rad/s, we can use the following conversion factor:
1 rev/min = (2π/60) rad/s
Now we can calculate the change in angular speed:
Δω = final angular speed - initial angular speed
= (60 rev/min) * (2π/60 rad/s) - 0
= 2π rad/s
Next, we can substitute the values into the equation:
100 J = 1/2 * I * (2π rad/s)^2
Simplifying the equation:
100 J = 1/2 * I * 4π^2 rad^2/s^2
200 J = I * 4π^2 rad^2/s^2
Now, solve for I:
I = 200 J / (4π^2 rad^2/s^2)
I ≈ 50.72 kg·m^2
Therefore, the moment of inertia of the flywheel is approximately 50.72 kg·m^2.