ToughSTEM
ToughSTEM
A question answer community on a mission
to share Solutions for all STEM major Problems.
Cant find a problem on ToughSTEM?
0
The combination of an applied force and a constant frictional force produces a constant total torque of 35.8 N·m on a wheel rotating about a fixed axis. The applied force acts for 6.04 s. During this time the angular speed of the wheel increases from 0 to 10.2 rad/s. The applied force is then removed, and the wheel comes to rest in 59.7 s
(a) Find the moment of inertia of the wheel.
_______ kg*m2
(b) Find the magnitude of the frictional torque.
_______ N*m
(c) Find the total number of revolutions of the wheel.
_______ rev
Edit
Community
1
Comment
Solutions
0
a)
Angular acceleration ? = (?2- ?1)/t = 10.2/6.04 = 1.688 rad/s^2

I ? = I *1.688 = 35.8

I = 35.8/ 1.688 = 21.208 kg m/s^2
===========================
b)
The wheel comes to rest only due to the frictional torque ? (friction)

? (friction) = I ?' = 21.208 ?'
?' = - 10.2/59.7 = - 0.170 minus to show that the speed reduces.
? (friction) = - 21.208*0.170 = 3.605 N.m
==============================
c)
in time 6.04s
?= 0.5 ? t^2 = 0.5*1.688*6.04^2 =30.79 radians.

Or from average angular velocity *time = (10.2/2)*6.04 =30.08 radians

In time 59.7 s

Total angle traversed
304.47 / (2?) revolutions = 48.45 revolutions
Edit
Community
1
Comment
Close

Choose An Image
or
Get image from URL
GO
Close
Back
Close
What URL would you like to link?
GO
α
β
γ
δ
ϵ
ε
η
ϑ
λ
μ
π
ρ
σ
τ
φ
ψ
ω
Γ
Δ
Θ
Λ
Π
Σ
Φ
Ω
Copied to Clipboard

to interact with the community. (That's part of how we ensure only good content gets on ToughSTEM)
OR
OR
ToughSTEM is completely free, and its staying that way. Students pay way too much already.
Almost done!