Calculating Torque for a Moving Vehicle Wheel: A Solid Disc Approach

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Homework Help Overview

The discussion revolves around calculating torque and analyzing energy conservation in the context of vehicle wheels and rolling cylinders. The original poster presents a problem involving a vehicle wheel treated as a solid disc and seeks to determine the torque applied as the vehicle accelerates. Another participant introduces a separate problem regarding a hollow cylinder rolling down a slope, prompting discussions on energy conservation principles.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore calculations related to torque and angular acceleration for the vehicle wheel, with some questioning the accuracy of their results. In the cylinder problem, there are discussions about using conservation of energy to find the angle of incline, with participants expressing uncertainty about their approach and assumptions regarding potential and kinetic energy.

Discussion Status

The discussion is active, with participants sharing their calculations and reasoning. Some guidance has been offered regarding the use of conservation of energy, and there is an ongoing exploration of the relationships between different physical quantities. Multiple interpretations and approaches are being considered, particularly in the context of the cylinder problem.

Contextual Notes

Participants are working under the constraints of homework assignments, which may limit the information available or the methods they can use. There is also a mention of an open-book test approaching, which may influence the urgency and depth of the discussion.

ar202
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Homework Statement



A vehicle wheel has a mass 2.5 kg and diameter 310 mm. The vehicle moves from rest to a linear velocity of 13 kmh -1 in 60 s. Assuming that the wheel can be treated as a solid disc, find the torque applied to the wheel

Homework Equations



Not so sure on how to add equations that aren't basic...

The Attempt at a Solution



I started with

0.5 x 2.5 x 0.1552 = 0.3003125

then

3.61ms-1 / 0.155 = 23.29 rads-1

23.29-0/60=0.388172043 (angular acceleration)

so then it should be

0.388172043 x 0.3003125 = 0.1165729167 Nm right? :confused:
 
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Your calculation appease to be correct.
 
i'm on my last question, i'll put it here instead of starting numerous threads... :rolleyes:

A hollow cylinder of mass 50 g and external and internal diameters of 50 mm and 44 mm respectively, rolls down a slope for a distance of 5 m without slipping. If the cylinder starts from rest and has a velocity of 3 ms-1 by the time it reaches the 5m point, what was the angle of incline of the slope

i'm a bit lost on where to start... i know it's something to do with potential energy...

btw thanks for looking at the other one
 
use conservation of energy.
 
benhou said:
use conservation of energy.

ke = pe ?

h = 1/2 x 32/9.81 ?

can i then assume it's a right angle triangle? or am i off target...? :redface:
 
Initial velocity, final velocity and displacement of the cm of the cylinder is given. Find the acceleration of the center of mass.
The equation of motion for translation a nd for rotation about the axis through cm is
ΣFx = Mg*sinθ + (-f) = M*a(cm) ...(1)
Στ = f*R = Iα = I*R*a(cm) ...(2) where f is the frictional force.
Substitute the value of I in the eq.2 and find the expression for a(cm). From that find θ.
 
ok went over some notes... was getting carried away

i've got

0.05x9.81x5Sinθ=0.225+0.198 (all sources of ke)

sinθ=0.1724 radians right?

180/pi x 0.1724 = 9.91 degrees?
 
ar202 said:
ke = pe ?

h = 1/2 x 32/9.81 ?

can i then assume it's a right angle triangle? or am i off target...? :redface:

Well, what is its total kinetic energy? By the way, it would be helpful if you shown all your work. I was a little lost for a few minutes.
 
benhou said:
Well, what is its total kinetic energy? By the way, it would be helpful if you shown all your work. I was a little lost for a few minutes.

yes apologies! bit lazy of me :-p
 
  • #10
ar202 said:
ok went over some notes... was getting carried away

i've got

0.05x9.81x5Sinθ=0.225+0.198 (all sources of ke)

sinθ=0.1724 radians right?

180/pi x 0.1724 = 9.91 degrees?

ok, so you used conservation of energy instead of forces. But could you explain how you got "0.225+0.198 (all sources of ke)". It seems too small for me.
 
  • #11
Never mind, I got it. It's the mass 0.05.
It looks good now.
 
  • #12
benhou said:
ok, so you used conservation of energy instead of forces. But could you explain how you got "0.225+0.198 (all sources of ke)". It seems too small for me.

1/2x0.05x32=0.225

1/2x1.1x10-4x602=0.198

i've done 3/0.05=60 rads-1 used the radius of the first diameter, no idea if this is correct or not or whether there is a way of using them both.
 
  • #13
I believe 1.1*104 is the moment of inertia, could you show me how you did it as well.
 
  • #14
ok cool, cheers. does my answer seem ok in your opinion?

having an open-book test in a couple of days :smile: thought i would do as many examples as possible for reference.
 
  • #15
benhou said:
I believe 1.1*104 is the moment of inertia, could you show me how you did it as well.

moment of inertia for a hollow disc = 1/2xm(r1^2+r2^2) if that makes sense. sorry I'm not familiar with how to add fancy text for formulas
 
  • #16
It's ok for now, but it would be really helpful if you did know. Here's how to learn latex code:
https://www.physicsforums.com/misc/howtolatex.pdf

moment of inertia for a hollow disc = 1/2xm(r1^2+r2^2) if that makes sense. sorry I'm not familiar with how to add fancy text for formulas"

I thought the + should be -. But ya, you are right. Good luck with your review!
 
  • #17
Is it possible to work out an objects angular velocity when you're only give it's moment of inertia, mass and radius...?
 

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