• Support PF! Buy your school textbooks, materials and every day products Here!

Please help me out with this question on torque

  • #1
208
0
Please help me out with this question on torque....

Homework Statement


A cylinder of mass M and radius R is resting on a horizontal platform(which is parallel to the x-y plane) with its axis fixed along the y -axis and free to rotate about its axis. The platform is given a motion in the x-direction given by x=Acos(wt). There is no slipping between the cylinder and platform.
Find The maximum torque acting on the cylinder during its motion?



Homework Equations



d2x/dt2=a

for no slipping a=R(alpha) alpha is dw/dt....



The Attempt at a Solution



I found maximum acceleration of platform and equated it as R times alpha.....
found alpha from there and then used torque = I*alpha

So max acceleration of platform is (neglecting sign) Aw2

alpha=Aw2/R

I=mR2/2

You get torque as =MRAw2/2

Answer is MRAw2/3

please reply....thank you.....
 
Last edited by a moderator:

Answers and Replies

  • #2
collinsmark
Homework Helper
Gold Member
2,894
1,232


Hello Abhishekdas,

For what it's worth, I also come up with
τ = AMRω2/2.​
Maybe it's a mistake in the book? (Unless maybe the cylinder isn't a solid, uniform cylinder, --i.e. unless maybe there's some information about the cylinder that's missing from the problem statement.)
 
  • #3
208
0


Ya even I think its a mistake.....

Anyway thanks a lot collinsmark.......I'm at least assured now my method is correct since you got the same answer.....
So I guess I wont worry about it right now.....
 
Last edited by a moderator:
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
250
Hi Abhishekdas! :smile:

(have an alpha: α and an omega: ω :wink:)
… for no slipping a=R(alpha) alpha is dw/dt....
noooo :redface:

that only applies to relative motion between the rolling surfaces …

you need to add the actual motion of the platform …

start again! :smile:

(and yes, I have got the same answer as the book! :wink:)
 
  • #5
collinsmark
Homework Helper
Gold Member
2,894
1,232


Hi Abhishekdas! :smile:

(have an alpha: α and an omega: ω :wink:)
Abhishekdas said:
for no slipping a=R(alpha) alpha is dw/dt....
noooo :redface:

that only applies to relative motion between the rolling surfaces …

you need to add the actual motion of the platform …

start again! :smile:

(and yes, I have got the same answer as the book! :wink:)
I've been thinking about this for awhile. Perhaps the discrepancy originates from how Abhishekdas and I are interpreting the set-up. (?) But the more I think about it, the more I think that we're interpreting it correctly.

Here's the way I imagine the setup, given the wording in the problem statement. I'll be using three dimensional Cartesian coordinates. z is the up/down direction, y is the left/right direction, and x is the forward/back direction. The crux of what I'll describe comes from the phrase in the first sentence, "...with its axis fixed along the y -axis...".

Imagine a rolling-pin (used for baking) which has handles connected by a small internal rod. Outside this rod is a wooden cylinder, sharing the same axis as the rod, which is free to rotate around the axis.

Now imagine that each handle is secured to the kitchen counter-top with a vise. The rolling-pin as a whole does not move at all (translationally), although the wooden cylinder can still rotate. The axis that the rolling pin is fixed to (via the vises) is the y-axis.

Sandwiched between the wooden cylinder and the counter-top is a flat breadboard. The breadboard is allowed to move forward or backward (along the x-axis), causing the wooden cylinder to rotate on its axis as the breadboard moves.

According to my calculations, such a setup would produce a maximum torque on the cylinder of τ = AMRω2/2, given the description of the breadboard's motion, as described in the problem statement.

==================

But is the above the right way to interpret the problem statement? Could it be that when the the problem statement says, "...with its axis fixed along the y -axis..." it's referring to the breadboard rather than the rolling-pin? (Such that the rolling-pin isn't secured to anything at all?) But that doesn't make much sense to me. That would imply that the breadboard has a preferred "axis." To me, the only way it makes sense is assume that the rolling-pin's central axis is the thing that is fixed in place (along the y-axis).

Are we missing something else?

[Edit: Abhishekdas, there wouldn't happen to be a figure or diagram that goes with this problem would there?]
 
Last edited:
  • #6
collinsmark
Homework Helper
Gold Member
2,894
1,232


Okay, I think I see what tiny-tim is getting at. :smile: But it requires that the problem be interpreted differently than what I had in my above post. (Which I still think that our previous interpretation is valid given the way the problem statement was worded). But anyway, setting up the situation differently...

If, instead of fixing the axis of the cylinder to the y-axis, let it roll freely on top of the platform, then the answer comes out to be what the book has.

In other words, although the cylinder's axis is parallel to the y-axis, it's not fixed to the y-axis.

Still in other words, remove the rolling-pin from the vise-clamps. That gives you the book's answer. And that actually makes for a more interesting problem too.

Abhishekdas, are you sure the problem was worded correctly in the original post's problem statement section? Because I still say, given how it was worded, it implies the axis of the cylinder is held in place on the y-axis. But maybe there was a mistake somewhere and the problem meant to say that the cylinder's axis wasn't fixed.
 
  • #7
208
0


hi....collinsmark.....

I really appreciate your effort on understanding ths problem.....
First thing there was no diagram in the book......And yes i checked my statement of the problem......Its same....

So basically you mean to say from what you described about the rolling pin example is that the axis is fixed along the y axis ie no motion of CM of the body right?....Which is what we have done i guess.....
So you mean to say that now......instead of fixing the axis just make sure the axis is aligned along the y axis and the CM can move right.....?

So thank you......and yes i will attempt this problem on the second idea and let you know.....
 
Last edited by a moderator:
  • #8
208
0


Hey collinsmark...you do get MRAw2/3 if you consider the axis of the cylinder to be free.......So maybe your approach is the right one.......hmmmm...thanks.....

And yes tiny-tim...Im afraid i didnt exactly get what you were trying to say.....We made a mistake that we considered the axis to be FIXED to the y axis....While we got the answer in the book by making it free......Is that the same thing you were trying to say? Could you please clarify yourself and yes if you were saying anything else i would be glad to know.....Thanks.....
 
Last edited by a moderator:
  • #9
tiny-tim
Science Advisor
Homework Helper
25,832
250
Hi Abhishekdas! :smile:
And yes tiny-tim...Im afraid i didnt exactly get what you were trying to say.....We made a mistake that we considered the axis to be FIXED to the y axis....While we got the answer in the book by making it free......Is that the same thing you were trying to say?
Yes. :smile:

It honestly never occurred to me that there was any other way of reading it …

if the axis was fixed, there'd be no point in the question saying "There is no slipping between the cylinder and platform" :rolleyes:
 
Last edited by a moderator:
  • #10
collinsmark
Homework Helper
Gold Member
2,894
1,232


Hey collinsmark...you do get MRAw2/3 if you consider the axis of the cylinder to be free.......So maybe your approach is the right one.......hmmmm...thanks.....
Yes, if the cylinder is free, I come up with τ = AMRω2/3 for the maximum torque.

And I have a hunch that a freely rolling cylinder is what the problem meant to imply (even if it wasn't worded clearly). If the cylinder is fixed, the problem is mostly just one of geometry, with only a single invocation of τ = Iα (and maybe a couple time derivatives) for the physics part. But if the cylinder is free to roll back and forth (such that its axis isn't fixed in space), the physics of the problem are quite a bit more interesting.

So, in retrospect, I'm guessing that the problem is just poorly worded, but you're supposed to assume that the cylinder is free to roll.
 
  • #11
208
0


Yes collinsmark....the physics part does become more intersting if the axis is free and i guess that is what the author meant.....

And yes tiny-tim....even if the axis is not fixed there could have been a case of slipping......If you think of a case where the coefficient of friction is lets say not enough to prevent slipping when the platform has maximum acceleration.....I think all you can infer from the no slipping part is you can go ahead assuming any value of friction to meet the condition for no slipping....thats what i think....what do you say???
 
  • #12
tiny-tim
Science Advisor
Homework Helper
25,832
250
And yes tiny-tim....even if the axis is not fixed …
Not following you :redface:

the axis is not fixed (except that it must stay parallel to the y-axis) :confused:
 
  • #13
208
0


Sooorrryyy......i meant.... 'even if the axis is fixed'.......now i think what i wrote makes sense......
 
Top