A torque of 77.7 Nm causes a wheel to start from rest, completes 5.55 revolutions and attains a final angular velocity of 88.8 rad/sec. What is the moment of inertia of the wheel?
I am trying to find angular acceleration first. How do I find it given this information? I need to know time...Also, I don't see how that information about revolutions will be used. What do I do?
Thank you!
the answer is 0.687
Convert 5.55 revolutions into radians. 1 rev = 2pi rad, so 5.55rev = 34.87 rad.
Use the equation ωf^2 = ωi^2 + 2α∆θ.
ωf = 88.8rad/sec, ωi = 0, and ∆θ = 34.87 rad.
Knowing those variables, you can calculate α = 113.069.
Finally, you can use the equation τ = Iα (where τ = torque, I = moment of inertia, and α = angular acceleration) to calculate the moment of inertia.
τ = 77.7Nm and α = 113.069.
I = τ/α = 77.7/113.069 = 0.687
Hope that helps
Well, it seems like you're in a bit of a torque-y situation, but don't worry, I'm here to help! To find the angular acceleration, you don't actually need the time or the information about the revolutions. Instead, you can use the torque!
The torque equation is given by:
τ = Iα
where τ is the torque, I is the moment of inertia, and α is the angular acceleration. In this case, the torque is 77.7 Nm.
Now, since the wheel starts from rest and attains a final angular velocity of 88.8 rad/sec, we can use the equation:
ω^2 = ω0^2 + 2αθ
where ω is the final angular velocity, ω0 is the initial angular velocity (which is 0 in this case since the wheel starts from rest), α is the angular acceleration, and θ is the angle through which the wheel rotates.
Since the initial angular velocity is 0, the equation simplifies to:
ω^2 = 2αθ
Substituting the values given, we get:
(88.8)^2 = 2α(5.55 * 2π)
Now you have one equation with one unknown (α). You can solve for α using simple algebra, and once you have the angular acceleration, you can use it to find the moment of inertia by rearranging the torque equation:
I = τ / α
Plug in the torque value (77.7 Nm) and the calculated angular acceleration, and voila! You'll have the moment of inertia of the wheel.
Good luck, and may the torque be with you!
To find the angular acceleration, we need to use the formula:
τ = I α
Where τ is the torque applied, I is the moment of inertia, and α is the angular acceleration.
In this case, we are given the torque (τ = 77.7 Nm) and we need to find the moment of inertia (I). To find α, we need to use the information about the wheel starting from rest and attaining a final angular velocity after completing a certain number of revolutions.
Let's break down the problem step-by-step:
1. First, convert the number of revolutions (5.55) into radians. Since one revolution is equal to 2π radians, multiply the number of revolutions by 2π:
θ = 5.55 rev * 2π rad/rev = 5.55 * 2π rad
2. Next, determine the final angular velocity (ωf) in radians per second. The angular velocity is given as 88.8 rad/sec.
3. Use the kinematic equation to find the initial angular velocity (ωi):
ωf^2 = ωi^2 + 2αθ
Since the wheel starts from rest (ωi = 0), we can simplify the equation to:
ωf^2 = 2αθ
Plug in the values:
(88.8 rad/sec)^2 = 2α(5.55 * 2π rad)
Solve for α:
α = (88.8^2) / (2 * 5.55 * 2π) rad/sec^2
4. Now that we have the angular acceleration (α), we can find the moment of inertia (I) using the torque equation:
τ = I α
Plug in the values:
77.7 Nm = I * (88.8^2) / (2 * 5.55 * 2π) rad/sec^2
Solve for I:
I = (77.7 Nm) * (2 * 5.55 * 2π) rad/sec^2 / (88.8^2)
This will give you the moment of inertia of the wheel.
torque = I alpha
where alpha = d omega/dt = angular acceleration
now
velocity proportional to time so
average angular velocity = 88.8/2 = 44.4 rad/s
does 5.55 *2pi radians
so
time = 5.55 * 2 pi /44.4 rad/s
= .0765 seconds to accelerate
acceleration = alpha = change in omega/ change in time
= 88.8/.0765 = 1161 rad/s^2
and of course
77.7 = I (1161)
do not trust my arithmetic.
time = 5.55 * 2 pi /44.4 rad/s
= .785 seconds to accelerate
acceleration = alpha = change in omega/ change in time
= 88.8/.785 = 113 rad/s^2
and of course
77.7 = I (113)