To tackle this problem, we need to break it down into manageable parts. We have a wheel experiencing a constant torque, which leads to a change in angular velocity, and then it experiences friction when the external torque is removed. Let’s go through each part step by step.
Calculating the Moment of Inertia
First, we need to find the moment of inertia (I) of the wheel. We can use the relationship between torque (τ), moment of inertia (I), and angular acceleration (α). The formula is:
τ = I * α
We know the torque (τ) is 20 N·m. To find angular acceleration (α), we need to convert the change in angular velocity from revolutions per minute (r.p.m) to radians per second (rad/s). The final angular velocity (ω) is given as 100 r.p.m:
- Convert 100 r.p.m to rad/s:
- 100 r.p.m = 100 * (2π rad / 1 min) = 100 * (2π rad / 60 s) = (100 * 2π) / 60 rad/s ≈ 10.47 rad/s
Since the initial angular velocity (ω₀) is 0 (the wheel starts from rest), we can find angular acceleration (α) using the formula:
α = (ω - ω₀) / t
Substituting the values:
α = (10.47 rad/s - 0) / 10 s = 1.047 rad/s²
Now we can substitute α back into the torque equation to find I:
20 N·m = I * 1.047 rad/s²
Solving for I gives:
I = 20 N·m / 1.047 rad/s² ≈ 19.1 kg·m²
Finding the Frictional Torque
Next, we need to determine the frictional torque (τ_f) that stops the wheel. When the external torque is removed, the wheel decelerates due to friction. We can use the same relationship between torque, moment of inertia, and angular acceleration:
τ_f = I * α_f
First, we need to find the angular deceleration (α_f). The wheel goes from 10.47 rad/s to 0 in 100 seconds, so:
α_f = (0 - 10.47 rad/s) / 100 s = -0.1047 rad/s²
Now substituting the values into the torque equation:
τ_f = 19.1 kg·m² * (-0.1047 rad/s²) ≈ -2.00 N·m
The negative sign indicates that this torque opposes the motion, which is expected for frictional torque.
Calculating the Total Revolutions
Finally, we want to find the total number of revolutions made by the wheel in 110 seconds. The first 10 seconds are under the influence of the external torque, and the next 100 seconds are under the influence of friction.
During the first 10 seconds, the wheel accelerates to 100 r.p.m, which we already calculated as 10.47 rad/s. The number of revolutions (N) during this time can be calculated as:
N = ω * t / (2π)
Substituting the values:
N = 10.47 rad/s * 10 s / (2π) ≈ 16.67 revolutions
During the next 100 seconds, the wheel decelerates uniformly from 10.47 rad/s to 0. The average angular velocity during this time is:
ω_avg = (10.47 rad/s + 0) / 2 = 5.235 rad/s
The number of revolutions during this deceleration phase is:
N_f = ω_avg * t / (2π)
Substituting the values:
N_f = 5.235 rad/s * 100 s / (2π) ≈ 83.33 revolutions
Now, adding both parts together gives the total number of revolutions:
Total revolutions = 16.67 + 83.33 = 100 revolutions
Summary of Results
- Moment of Inertia (I): 19.1 kg·m²
- Frictional Torque (τ_f): 2.00 N·m
- Total Revolutions in 110 seconds: 100 revolutions
