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?
Time required to stop:
theta = (1/2)(w0 + w)*t
163(2π) rad = (1/2)(30(2π) rad/s +0 rad/s)*t
t ~ 10.87 s
Angular acceleration of wheel:
= (0 rad/s - 30*2π rad/s) / 10.87 s
~ -17.3 rad/s^2
Required slowing torque:
T = I * alpha
= (0.015 kg*m^2)(-17.3 rad/s^2)
~ -0.26 kg*m^2/s^2
(or -0.26 N*m)
Well, let me first compliment the wheel on its endurance. Going for 163 revolutions before deciding to take a little break is quite impressive. Now, to calculate the torque that slowed it down, we can use the equation:
Torque = Moment of Inertia * Angular Acceleration
Since the wheel starts at 30 rev/s and ends at rest, we can find the angular acceleration using the formula:
Final Angular Velocity^2 = Initial Angular Velocity^2 + 2 * Angular Acceleration * Number of Revolutions * 2π
Plugging in the values, we get:
0^2 = (30 rev/s)^2 + 2 * Angular Acceleration * 163 rev * 2π
Now, solving for the angular acceleration, we find:
Angular Acceleration = [0^2 - (30 rev/s)^2] / [2 * 163 rev * 2π]
Angular Acceleration = -0.00214 rev/s^2
Finally, we can calculate the torque:
Torque = Moment of Inertia * Angular Acceleration
Torque = 0.015 kg m^2 * (-0.00214 rev/s^2)
Torque = -0.0000321 kg m^2/s^2
So, the torque that slowed down the wheel is approximately -0.0000321 kg m^2/s^2. And yes, the negative sign is just the wheel's way of saying "I'm taking a break, don't bother me."
To find the torque that slowed the wheel, we can use the relationship between torque, moment of inertia, and angular acceleration.
First, let's calculate the final angular velocity (ωf) of the wheel when it comes to rest. We know that the wheel starts at an initial angular velocity (ωi) of 30 rev/s and stops after 163 revolutions.
Since the wheel has completed 163 revolutions, we can calculate the final angular position (θf) using the formula:
θf = 2π * number of revolutions
θf = 2π * 163
θf = 1024π rad
Next, we can calculate the final angular velocity (ωf) using the formula:
ωf² = ωi² + 2αθ
ωf² = 0 + 2α * 1024π
Since the wheel comes to rest, the final angular velocity (ωf) is 0, so we can rearrange the equation:
0 = 2α * 1024π
α = 0 rad/s²
Here, α represents the angular acceleration. Since the wheel is slowing down, the angular acceleration is in the opposite direction of the initial angular velocity.
Using the equation τ = Iα, where τ is torque, I is the moment of inertia, and α is the angular acceleration, we can now find the torque that slowed the wheel.
τ = Iα
τ = (0.015 kg m^2) * (0 rad/s²)
τ = 0 Nm
Therefore, the torque that slowed the wheel is 0 Nm.
To find the torque that slowed down the wheel, we can use the principle of conservation of angular momentum.
Angular momentum (L) is given by the product of the moment of inertia (I) and the angular velocity (ω). It can be expressed as L = Iω.
Given:
Mass of the wheel (m) = 0.500 kg
Moment of inertia (I) = 0.015 kg m^2
Initial angular velocity (ω_initial) = 30 rev/s
Final angular velocity (ω_final) = 0 rev/s (since the wheel is at rest)
First, let's convert the initial angular velocity from revolutions per second to radians per second. Since 1 revolution = 2π radians:
ω_initial = 30 rev/s * 2π rad/1 rev ≈ 188.5 rad/s
Using the conservation of angular momentum, we can equate the initial and final angular momentum:
L_initial = L_final
I * ω_initial = I * ω_final
Substituting the given values:
(0.015 kg m^2) * (188.5 rad/s) = (0.015 kg m^2) * ω_final
Simplifying the equation:
ω_final = (0.015 kg m^2 * 188.5 rad/s) / (0.015 kg m^2)
ω_final ≈ 188.5 rad/s
Since the wheel comes to rest after 163 revolutions, the change in angular velocity is:
Δω = ω_initial - ω_final
Δω = 188.5 rad/s - 0 rad/s
Δω = 188.5 rad/s
Finally, we can find the torque (τ) that slowed down the wheel using the equation:
τ = I * Δω
τ = (0.015 kg m^2) * (188.5 rad/s)
τ ≈ 2.828 N m
Therefore, the torque that slowed down the wheel is approximately 2.828 N m.