A 0.500 kg wheel that has a moment of inertia of 0.015 kg m^2 is initially turning at 30 rev/s. it starts to rest after 163 revolution. how large is the torque that slowed it?
wf=wi+2(angular acceleration)*angle
0=(30^2PI)^2+ 2*alpha*163*2PI
solve for alpha, angular acceleration.
Then
Torque=alpha*momentinertia
solve for torque
To determine the torque that slowed down the wheel, we first need to find the angular deceleration. We can do this using the equation:
Δω = (vf - vi) / t
where:
Δω is the change in angular velocity,
vf is the final angular velocity,
vi is the initial angular velocity,
and t is the time.
Given that the initial angular velocity, vi, is 30 rev/s and the final angular velocity, vf, is 0 rev/s (as the wheel comes to rest), we can substitute these values into the equation:
Δω = (0 rev/s - 30 rev/s) / 163 rev = -0.183 rev/s²
Since we want the angular deceleration in units of rad/s², we need to convert revolutions to radians. There are 2π radians in one revolution, so:
Δω = -0.183 rev/s² * 2π rad/rev ≈ -1.149 rad/s²
Now, we can calculate the torque using the formula:
τ = I * α
where:
τ is the torque,
I is the moment of inertia, and
α is the angular acceleration.
Given that the moment of inertia, I, is 0.015 kg m² and the angular acceleration, α, is -1.149 rad/s², we can substitute these values into the equation:
τ = 0.015 kg m² * -1.149 rad/s² ≈ -0.017 m² kg/s²
Therefore, the torque that slowed down the wheel is approximately -0.017 m² kg/s². Note that the negative sign indicates that the direction of the torque is opposite to the initial angular velocity of the wheel.
To determine the torque that slowed down the wheel, we can use the equation:
Torque = Change in Angular Momentum / Time
First, we need to find the change in angular momentum. The initial angular momentum (L1) can be calculated using the equation:
L1 = Moment of Inertia * Angular Velocity
Given:
Moment of Inertia (I) = 0.015 kg·m²
Angular Velocity (ω) = 30 rev/s
Time (t) = 163 revolutions
Converting angular velocity to radians per second:
ω = (30 rev/s) * 2π rad/rev = 60π rad/s
Substituting the values into the equation, we get:
L1 = (0.015 kg·m²) * (60π rad/s) = 0.9π kg·m²/s
Next, let's find the final angular momentum (L2). Since the wheel is at rest, the final angular velocity is 0 (ω = 0). Therefore, L2 = (0.015 kg·m²) * 0 = 0 kg·m²/s
The change in angular momentum (ΔL) is then given by:
ΔL = L2 - L1
ΔL = 0 kg·m²/s - 0.9π kg·m²/s = -0.9π kg·m²/s
Finally, we need to convert the revolutions to seconds.
Time (t) = 163 revolutions / 30 rev/s = 5.433 seconds
Now, substituting the values into the torque equation, we can solve for the torque:
Torque = ΔL / t
= (-0.9π kg·m²/s) / 5.433 s
Evaluating the expression, the torque that slowed down the wheel is approximately:
Torque = -0.165 N·m (rounded to three decimal places)
Note that the negative sign indicates that the torque opposes the initial angular momentum, causing it to slow down.