A combination of an applied force and a frictional force produce a constant total torque of 38 N-m on a wheel rotating about a fixed axis. This combined force acts for 8 s, during which the wheel’s angular speed increases from 0 to 12 rad/s. Find the moment of inertia of the wheel.
torque = I * change in omega/change in time
38 = I * (12/8)
I = 2 * 38/3
To find the moment of inertia of the wheel, we can use the equation for torque:
Torque (τ) = Moment of Inertia (I) × Angular Acceleration (α)
In the given problem, we have the value of torque (τ) as 38 N-m and the angular speed (ω) as 12 rad/s. We also know that the wheel starts from rest, so the initial angular speed (ω₀) is 0 rad/s. We need to find the moment of inertia (I).
We can find the angular acceleration (α) using the following equation:
Angular Acceleration (α) = (Final Angular Speed - Initial Angular Speed) / Time
Given:
Final Angular Speed (ω) = 12 rad/s
Initial Angular Speed (ω₀) = 0 rad/s
Time (t) = 8 s
Substituting these values into the equation, we can calculate the angular acceleration:
α = (12 - 0) / 8
α = 1.5 rad/s²
Now, we have the torque (τ), angular acceleration (α), and the formula for torque. Rearranging the formula, we can solve for the moment of inertia (I):
I = τ / α
I = 38 N-m / 1.5 rad/s²
I = 25.33 kg·m²
Therefore, the moment of inertia of the wheel is approximately 25.33 kg·m².