noospace
Posts: 46
Joined: Fri Feb 22, 2008 9:14 pm

3. A uniform 2 kg cylinder rests on a lab cart as shown. The coefficient of static friction between the cylinder and the cart is 0.5. If the cylinder is 4 cm in diameter and 10 cm high, which of the following is closest to the minimum acceleration required to cause the cylinder to tip over?

This question had me a bit confused for a while. I found that I could get the right answer by assuming the static friction is essentially infinite and then transforming into the cart frame where the cylinder experiences an inertial force of m*a in addition to gravity m*g acting down. Then applying torque balance about the lower corner of the cylinder gives me

5 cm * m *a = 2 cm * m * g

and thus a = 0.4* g = 4 m/s/s

Does anyone know of a more elegant approach to this question?

Thanks

blackcat007
Posts: 378
Joined: Wed Mar 26, 2008 9:14 am

### Re: Physics Olympiad 1999 Q3

3. A uniform 2 kg cylinder rests on a lab cart as shown. The coefficient of static friction between the cylinder and the cart is 0.5. If the cylinder is 4 cm in diameter and 10 cm high, which of the following is closest to the minimum acceleration required to cause the cylinder to tip over?

This question had me a bit confused for a while. I found that I could get the right answer by assuming the static friction is essentially infinite and then transforming into the cart frame where the cylinder experiences an inertial force of m*a in addition to gravity m*g acting down. Then applying torque balance about the lower corner of the cylinder gives me

5 cm * m *a = 2 cm * m * g

and thus a = 0.4* g = 4 m/s/s

Does anyone know of a more elegant approach to this question?

Thanks
moment balancing is the only condition for an eqm. thus i think this is the only pith of the problem..

noospace
Posts: 46
Joined: Fri Feb 22, 2008 9:14 pm

### Re: Physics Olympiad 1999 Q3

Yeah, but one would think that there's a better way than considering the moments of fictitious forces.

WhoaNonstop
Posts: 853
Joined: Mon Sep 21, 2009 1:31 am

### Re: Physics Olympiad 1999 Q3

Well, you can't really "exactly" solve the problem, and by viewing the other problems on the test, it appears that there isn't can't be much to it.

How I would go about "reasoning" the answer would be through F = ma.

Where, the force on the cylinder, the friction is uFn = ma.
Also Fn - mg = 0 (Fn = mg) and therefore ug = a.

Solving of course, a = 4.9 m/s^2.

However, since a force is exerted on the top end of the cylinder, the drag force, you could safely assume that it would fall over at less than 4.9 m/s^2. So 4.0 m/s^2 becomes the most reasonable choice.

I am not sure of an exact way to do it myself. It seems like there would have to be more info. =)

-Riley

noospace
Posts: 46
Joined: Fri Feb 22, 2008 9:14 pm

### Re: Physics Olympiad 1999 Q3

Hi WhoaNonstop,

No. I think this is a torque balancing problem as stated earlier.

michael
Posts: 50
Joined: Fri Sep 05, 2008 7:21 am

### Re: Physics Olympiad 1999 Q3

Hey,
I agree that the solution is not obvious to this problem. But I think the following method is fairly rigorous:

Consider moments about the COM of the cylinder - (correct me if I'm wrong, but we can always choose any point we like to take moments about -right?)

There are 3 forces on the block:
1) mg going down through the COM
2) a reaction force R, which when the cylinder is just about to tip, is upward, and equal to mg THROUGH THE BACK CORNER OF THE BLOCK, NOT THROUGH THE COM
3) a frictional force (F) in the direction of motion of the cart on the base of the cylinder

The maximum friction F_max that can be supplied to the cylinder is mu R = mu mg. For accelerations a, the force F will be:
m a for m a < F_max
F_max for m a > F_max

The first thing to check is that the max force is large enough to tip the cylinder (height h and diameter d):
m mu g (h/2) > mg (d/2) is equivalent to mu h > d which is true since 0.5* 10 = 5 > 4 so the block can tip and (E) is ruled out

Now check what min force causes tipping. When just about to tip, the condition of moment balance occurs:
m a_tip (h/2) = m g (d/2) thus a_tip = g(d/h) = 4m/s^2

I think the most dodgy bit of this argument is that mu would be different when all the force was acting through a point (i.e. just before tipping) than when spread over the surface. But I doubt they would consider those sorts of effects in this kind of a multiple choice test!