ω₀=2πn₀=2π•2000/60 rad/s
ω=2πn =2π•150/60 rad/s
ω=ω₀-εt
ε= (ω₀-ω)/t=….
L₀=I ω₀=…
L=I ω=…
L(ave) =(L+Lâ‚€)/2
Determine the angular acceleration.
What is angular momentum of the electric motor if the rotational inertia is 18.kg.square meter in initial velocity?
In final velocity?
What is the average angular momentum?
ω=2πn =2π•150/60 rad/s
ω=ω₀-εt
ε= (ω₀-ω)/t=….
L₀=I ω₀=…
L=I ω=…
L(ave) =(L+Lâ‚€)/2
angular acceleration (α) = (final angular velocity - initial angular velocity) / time
Given:
Initial angular velocity (ω1) = 2000 rpm
Final angular velocity (ω2) = 150 rpm
Time (t) = 3 sec
First, let's convert the angular velocities from rpm to rad/s:
ω1 = 2000 rpm * (2π rad/1 min) * (1 min/60 sec) = 2000 * 2π / 60 ≈ 209.44 rad/s
ω2 = 150 rpm * (2π rad/1 min) * (1 min/60 sec) = 150 * 2π / 60 ≈ 15.71 rad/s
Now we can calculate the angular acceleration:
α = (ω2 - ω1) / t
= (15.71 rad/s - 209.44 rad/s) / 3 sec
≈ -64.91 rad/s^2
Therefore, the angular acceleration is approximately -64.91 rad/s^2.
Next, let's determine the angular momentum of the electric motor at the initial and final velocities.
The formula for angular momentum (L) is:
L = rotational inertia (I) * angular velocity (ω)
Given:
Rotational inertia (I) = 18 kg·m^2 (in both cases)
For the initial velocity:
Angular velocity (ω1) = 209.44 rad/s (from earlier calculations)
L1 = I * ω1
= 18 kg·m^2 * 209.44 rad/s
≈ 3770.92 kg·m^2/s
For the final velocity:
Angular velocity (ω2) = 15.71 rad/s (from earlier calculations)
L2 = I * ω2
= 18 kg·m^2 * 15.71 rad/s
≈ 282.78 kg·m^2/s
Therefore, the angular momentum at the initial velocity is approximately 3770.92 kg·m^2/s, and at the final velocity is approximately 282.78 kg·m^2/s.
To calculate the average angular momentum, we can use the formula:
Average angular momentum = (Initial angular momentum + Final angular momentum) / 2
Average angular momentum = (3770.92 kg·m^2/s + 282.78 kg·m^2/s) / 2
= 2026.85 kg·m^2/s
Therefore, the average angular momentum is approximately 2026.85 kg·m^2/s.