# Inclined plane - a small sphere, a big sphere and a cylinder

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1. May 1, 2017

### ChessEnthusiast

1. The problem statement, all variables and given/known data
We place a cylinder with radius R, a sphere with radius R and a sphere with radius (2R). Which of the solids will roll out of the plane first? Assume no rolling resistance and no air resistance.

2. Relevant equations
1) The net torque is the torque of the force of static friction.
$$M_{bigSphere} = fm_{1}g\cos(\theta)$$
$$M_{smallSphere} = fm_{2}g\cos(\theta)$$
$$M_{cylinder} = fm_{3}g\cos(\theta)$$

2) The moments of inertia:
$$I_{bigSphere} = \frac{2}{5}m_{1}(4R^2)$$
$$I_{smallSphere} = \frac{2}{5}m_{2}R^2$$
$$I_{cylinder} = \frac{1}{2}m_{3}R^2$$

3. The attempt at a solution
1) I calculate the angular acceleration for each of the solids:
$$\epsilon_{bigSphere} = \frac{fm_{1}g\cos(\theta)}{\frac{2}{5}m_{1}(4R^2)} = \frac{5gf\cos(\theta)}{8R^2}$$
$$\epsilon_{smallSphere} = \frac{ fm_{2}g\cos(\theta)}{\frac{2}{5}m_{2}R^2} = \frac{5gf\cos(\theta)}{2R^2}$$
$$\epsilon_{cylinder} = \frac{2fg\cos(\theta)}{R^2}$$

Since the small sphere has the biggest angular acceleration, it will take it the least time to roll out of the plane.
Is my solution to this problem valid?

2. May 1, 2017

### haruspex

Does a small sphere travel as far per revolution as a larger sphere?

3. May 1, 2017

### kuruman

Torque about what point? Your expressions predict that when the objects are on a horizontal plane (θ = 0) the torque is non-zero. Does that make sense? Also, how do you know that the force of static friction acting on each object is the same?

4. May 2, 2017

### ChessEnthusiast

@kuruman
Right, my expression of the torque is not correct - it should have contained the radius as well and I should have used sine instead.
$$M = fm_{object}g\sin(\theta)R_{object}$$
This will result in one R cancelling out.

Apart from that, I should have used the linear acceleration instead of the angular one, which makes me arrive at the conclusion that both spheres will need the exactly same amount of time to roll out of the plane.

f is the static friction coefficient.

Last edited: May 2, 2017
5. May 2, 2017

### haruspex

Nearly right.
$$M =F_fR_{object}$$
where Ff is the frictional force, but that will only equal $\mu_sm_{object}g\sin(\theta)$ in the limiting case, where it is on the point of slipping.

6. May 2, 2017

### kuruman

Please explain what the coefficient of static friction has to do with anything. It comes into play only when two surfaces in contact are about to start sliding relative to one another. If the surfaces are not on the verge of sliding past each other (and there is no language in the problem that suggests that this is the case), the static friction is whatever is necessary to provide the observed acceleration.

Also, you did not clarify about what point you express the torque due to static friction. If it is about the pont of contact, this torque is zero (no lever arm); if it is about the CM, static friction is perpendicular to the lever arm, so $\sin \theta =1$.

7. May 2, 2017

### ChessEnthusiast

I assumed that the static friction is what causes the object to rotate. Should I have taken some other force of friction into account?
As for the the point about which I expressed the torque, I calculated it relatively to the CM of the object, hence - right, the sine will equal one.

8. May 3, 2017

### kuruman

There is no other friction present. However, the force of static friction acting on each object does not have the same value, therefore you cannot use the same symbol $f$ for all of them. So, if you wish to go this route, you need to find the force of friction on each object using the FBD method and Newton's 2nd law for both linear and rotational motion. An alternative that I would recommend is energy conservation.
That is correct.

9. May 3, 2017

### ChessEnthusiast

What about choosing the axis of rotation in the point of contact of the object and the plane?
Then, there I would have to take into consideration the "drag force" which is a component of the force of gravity, and the force of friction would cancel out
then, the torque:
$$M = Rmg\sin(\theta)$$

10. May 3, 2017

### kuruman

That's correct for the torque about the point of contact. Now what?

11. May 3, 2017

### ChessEnthusiast

Let's analyze the case of the cylinder.
$$M = Rmg\sin(\theta)$$
Moment of inertia:
$$I = \frac{1}{2}mR^2 + mR^2 = \frac{3}{2}mR^2$$
$$a = \frac{M}{I}R$$

12. May 3, 2017

### kuruman

That's the linear acceleration. Can you simplify the expression using what you have for $I$ and $M$?

13. May 3, 2017

### ChessEnthusiast

$$a = \frac{2}{3}gsin(\theta)$$
However, I'm worried that the acceleration does not depend on the friction coefficient.

14. May 3, 2017

### kuruman

As long as the rolling is without slipping the coefficient of static friction is irrelevant. The cylinder would take the same amount of time to reach bottom on a wooden incline as on a rubber-coated incline. It might be a good exercise for you to figure out the force of static friction on the cylinder. All you have to do is solve$F_{net}=ma$ in which you already know $a$.