Dynamics question (2 discs rolling down hill with different inertia)

[qwerty]

Hi guys

Haven't posted in this forum for a long long time.. good to read over some more interesting stuff on here i havent had the time to for a long long time.. anyway here goes

 

Dynamics (mechanical engineering) gave me this assignment question which is more of a study question for our final exam (because it doesn't count towards our marks)

 

Here is the question:

A thin ring and a circular disc, each of mass 'm' and radius 'R', are released from rest on an inclined surface and allowed to roll a distance 'D'. Determine the ratio of the times required.

 

I firstly drew my FBD (free body diagram) and this is it:

(take a close look at the direction of my frictional force, my tutor said its the other way (to the right) but the textbook says its the way i've drawn it, can someone tell me if i've done it correctly).

 

Now, I know the mass moment of inertia for each object,

SOLID DISC: I = 0.5 MR^2

THIN RING: I = MR^2

 

Now, I know that M = I . alpha

that is (sum of moments) = (inertia) x (angular accel)

 

So i know the inertia, I can work out sum of moments by looking at the FBD and the only force that would create a moment about the centre would be the frictional force. So sum of moments = -Ff x R.

 

Then I can find alpha (angular acceleration)..

 

Now I thought I can find the straight line acceleration of the centre of mass by using other equations but I dont know how to?

 

Then i might be able to find the ratio of times required?

 

_________________

 

I'm stuck on these last 2 lines... what do i do... what do i do once i've found alpha?

 

ALSO, I'm not sure if what i've done is correct... it seemed pretty logical to me but then again i could have easily done it wrong..

 

Thank you all very much i appreciate it a lot..

[swansont]

The angular and linear relations will be related by the radius of the wheels. [math]v = \omega r[/math] and [math]a = \alpha r[/math]

After that it should just be kinematics

[Royston]

(take a close look at the direction of my frictional force, my tutor said its the other way (to the right) but the textbook says its the way i've drawn it, can someone tell me if i've done it correctly)

 

For a wheel going downhill the friction should be in the same direction as the motion, i.e your tutor is correct. If the wheel was going uphill then the direction of the frictional force would be opposite to the direction of motion.

[uncool]

For a wheel going downhill the friction should be in the same direction as the motion, i.e your tutor is correct. If the wheel was going uphill then the direction of the frictional force would be opposite to the direction of motion.

 

Shouldn't it be preventing the motion, i.e. going against the motion?

=Uncool-

[Royston]

Shouldn't it be preventing the motion, i.e. going against the motion?

=Uncool-

 

I had a similar problem to this last year, so that was just from memory...I found this, but this could be wrong as well.

 

http://webphysics.davidson.edu/faculty/dmb/PY430/Friction/rolling.html

Note that the friction can be in the direction of motion (rolling downhill) or opposite to it (rolling uphill). In pure rolling motion there is no sliding or slipping, thus the contact points have no relative motion (no relative velocity). This results in a frictional force of zero. Therefore, the wheel will roll forward with constant velocity, v = Rw, where R is the radius of the wheel.

 

 

But yeah, friction is resistance so obviously it's an opposing force to the direction of motion, perhaps somebody can clarify on downhill motion.

[qwerty]

So, basically the frictional force acts the opposite direction to which the wheel would be accelerating (up the hill, wheel accelerates to the right at the point of contact - if the wheel was rolling to the right), therefore Ff would be to the left at tthe point of contact.

 

in my case, wheel is accelerating down hte hill and at point of contact that point of the wheel would be accelerating to the left, so the Ff is to the right..

 

My textbook is still wrong in this case... unless im not understanding something..

 

______________

 

Back on topic..

I solved a = (alpha) R for each case, and worked out a = Ff/m and 2Ff/m for the ring and disc respectively.

 

Then integrated with respect to time to get velocity... the nagain to get displacement.

 

Subbed in when s = D.. t = sqrt(2Dm/Ff) and sqrt(Dm/Ff) for the ring and disc.

 

Dividing the two to get the ratio gave me Tring/Tdisc = sqrt(2) : 1

 

This is the wrong answer , the answer should be Tring/Tdisc = Sqrt(4/3)

 

I'll try again with the frictional force in the opposite direction see if that gives me the right answer..

 

can anyone notice anything i've done absolutely wrong..

 

Thank you very much.

 

found it..

 

went to see tutor today again and this one actually helped me..

 

Thanks for your help guys

 

also this other textbook i went to get shows the friction in the same way the other textbook had it..

 

I have no idea what to do wtih friction. The tutor said it doesn't matter what you assume, the answer will tell you if its right or wrong so i suppose it doesn't matter?

 

Cheers all!!

[Fred56]

My Physics 101 textbook says that the frictional force "always opposes the motion" of a moving body, uphill, downhill, or no hill. Then goes on to give two examples, one uphill, one downhill, and the friction vector points in the opposite direction to the direction of motion. The example with no slope also does this. But then the examples are all for brick-style bodies rather than wheels with only one point of contact.

 

That last sentence might be a clue to why your tutor says it doesn't matter.

[swansont]

I think it's more straightforward to attack this from an energy standpoint. Since the acceleration is constant, the final speed is proportional to the time.

 

KE_{translational} + KE_rotational = PE_initial

 

Solve for v or [math]\omega[/math] for each case, and take the ratio.

[Fred56]

There certainly is a certain elegance to that. Kinematics is "energy in motion".

 

I just though of a way to illustrate another idea about kinematics. If a wheel or a ring rolls down an inclined surface, any point of contact is like a time derivative of the body's motion. If you set up an experiment with an inclined surface that made a regular sequence of marks on the moving diameter you could show this to anyone. Neat or what?

 

Ahem.

-except they would need to be regular intervals of time. After thinking on this, I have decided that this would present some difficulty. The aim is to produce a series of marks on the outer edge of a rolling body, and to do this in a timed way...hmmm

[WintersGirl]

can u help me for a moment?

 

please? anyone?

[Fred56]

Um, what's up?

[theCPE]

Shouldn't it be preventing the motion, i.e. going against the motion?

=Uncool-

 

The OP's FBD is correct. The frictional force should point uphill. In this incident the friction helps (enables) angular acceleration and thus must be opposite the disc's motion and with the rotational motion.

 

Imagine if there was no friction the disc would slide, right?

 

Now instead of lets say you were talking about the driving wheels of a vehicle the friction is with the direction of the vehicle. If there was zero friction the vehicle would not go anywhere and the wheels would just spin, the friction pushes the vehicle forward.