To tackle this problem, we need to break it down into manageable parts. We have a wheel subjected to a constant torque, and we need to find its moment of inertia, the frictional torque, and the total revolutions made during a specific time frame. Let's go through each part step by step.
1. Finding the Moment of Inertia
The moment of inertia (I) can be calculated using the relationship between torque (τ), moment of inertia, and angular acceleration (α). The formula is:
τ = I * α
First, we need to determine the angular acceleration. We know the initial angular velocity (ω₀) is 0, the final angular velocity (ω) is 100 revolutions per minute (r.p.m.), and the time (t) is 10 seconds. We can convert the final angular velocity to radians per second:
ω = 100 r.p.m. × (2π rad / 1 rev) × (1 min / 60 s) = (100 × 2π / 60) rad/s ≈ 10.47 rad/s
Now, we can find the angular acceleration using the formula:
α = (ω - ω₀) / t = (10.47 rad/s - 0) / 10 s = 1.047 rad/s²
Now substituting the values into the torque equation:
20 N·m = I * 1.047 rad/s²
Solving for I gives:
I = 20 N·m / 1.047 rad/s² ≈ 19.1 kg·m²
2. Determining the Frictional Torque
Once the external torque is removed, the wheel is stopped by friction over 100 seconds. The frictional torque (τ_f) can be calculated using the same relationship:
τ_f = I * α_f
Here, α_f is the angular deceleration. Since the wheel comes to a stop, the final angular velocity is 0. We can find α_f using:
α_f = (ω - 0) / t = (10.47 rad/s - 0) / 100 s = 0.1047 rad/s²
Now substituting into the frictional torque equation:
τ_f = 19.1 kg·m² * 0.1047 rad/s² ≈ 2.0 N·m
3. Calculating the Total Revolutions in 110 Seconds
To find the total revolutions made by the wheel in 110 seconds, we need to consider both the time under torque and the time under friction. The wheel accelerates for 10 seconds and then decelerates for 100 seconds.
Revolutions during Acceleration
The angular displacement (θ) during acceleration can be calculated using:
θ = ω₀ * t + 0.5 * α * t²
Substituting the known values:
θ = 0 * 10 + 0.5 * 1.047 rad/s² * (10 s)² = 0.5 * 1.047 * 100 = 52.35 rad
Revolutions during Deceleration
During deceleration, we can use the same formula, but we need to find the angular displacement while the wheel is slowing down:
θ_f = ω * t - 0.5 * α_f * t²
Substituting the values:
θ_f = 10.47 rad/s * 100 s - 0.5 * 0.1047 rad/s² * (100 s)²
θ_f = 1047 rad - 0.5 * 0.1047 * 10000 = 1047 rad - 523.5 rad = 523.5 rad
Total Revolutions
Now, we can find the total angular displacement:
θ_total = θ + θ_f = 52.35 rad + 523.5 rad = 575.85 rad
To convert this to revolutions:
Revolutions = θ_total / (2π) ≈ 575.85 rad / (2π) ≈ 91.7 revolutions
Summary of Results
- Moment of Inertia (I): 19.1 kg·m²
- Frictional Torque (τ_f): 2.0 N·m
- Total Revolutions in 110 seconds: Approximately 91.7 revolutions
This breakdown should clarify how we arrived at each answer and the relationships between torque, moment of inertia, and angular motion. If you have any further questions or need clarification on any part, feel free to ask!
