Sunday, November 5 11:12 PM f1234567890@mail.fi FlipltPhysics macmillan learning
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Sunday, November 5 11:12 PM f1234567890@mail.fi FlipltPhysics macmillan learning Physics 4A: Mechanics and Wave Motion, F2017 California State University, Fresno Unit 18: Prelecture / Checkpoint Honework Homework: HW 10 Of Ch9 Deadline: 100% until Tuesday, November 7 at 11:59 PM Tipler6 9.P.060. A wheel free to rotate about its axis that is not frictionless is initially at rest. A constant external torque of-51 N-m is applied to the wheel for 19 s. giving the wheel an angular velacity of -552 rev/min. The external torque is then removed, and the wheel comes to rest 120 s later. (Include the sign in your answers.) 1) (a) Find the moment of inertia of the wheel. kg- m2 Submt You currently have 0 submissions for this question, Only 1O submission are allowed. You can moke 10 more submissions for this question. 2) b) Find the frictional torque, which is assumed to be constant. N·m) Submt You currently have 1 submissions for this question. Only 10 submission are allowed. Hou con make 9 more subrnissions for this question Copyright & 2017 Freenan Worth Publishers a diision of Macmillan Learning About I Tech Support i Find Your Lecal Sales Rep Terms Of Use 1 Prvacy PelcyExplanation / Answer
Torque = F.d = I.a
I = moment of inertia
a = angular acceleration
F = force
d = distance.
The first torque made the wheel rotate from the rest until the speed gained an angular velocity of 552 rev / min
The torque is 51 Nm
51 Nm = I*a
I = moment of inertia of the wheel
a = angular acceleration
let's find a, with : Wf = 552 rev / min, time = 19 s, initial velocity = 0 m/s
552*2pi / 60 = a*19
a = 3.04 rad/s^2
then : 51 = I*3.04 >>> I = 16.8 kg*m^2
b), Let's find the frictional torque with :
Torque = 16.8*a'
a' = final angular acceleration ( when the wheel stops)
initial speed before the braking : 552*2pi / 60 rad / s
final velocity = 0 rad / s, time = 120 s
0 = 552*2pi/60 - a'120
a' = 0.48 rad / s^2
Torque = 16.8*0.48 = 8.064 Nm
Hope that helped……!!!
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