Is this formula for a ball rolling down a ramp incorrect?

[etotheipi]

Homework Statement

See below

Relevant Equations

N/A

I've got to do an experiment that essentially involves rolling a ball bearing down a (frictional) ramp and measuring its acceleration. It's quoted in the manual that the linear acceleration of a ball bearing rolling down a ramp at angle $\theta$ is $a = \frac{5}{9} g \sin{\theta}$. When I worked it out myself, I wrote down two equations$$mg\sin{\theta} - F = ma$$ $$rF = \frac{Ia}{r} \implies F = \frac{Ia}{r^2}$$which then gives$$mg\sin{\theta} = a \left(m + \frac{I}{r^2} \right) \implies a = \frac{g\sin{\theta}}{1 + \frac{I}{mr^2}}$$For a ball bearing, a sphere, $I = \frac{2}{5} mr^2$, so we should get$$a = \frac{5}{7} g\sin{\theta}$$which is a different coefficient to the one quoted. I thought it was pretty unlikely that there's a mistake in the handbook, so I presumed there was an aspect of the model that I haven't included in the analysis above. I wondered if anyone could tell how to get to the quoted equation? Thanks!

 

[haruspex]

I assume it is solud and uniform.
5/7 sounds very familiar and your analysis is correct. Never seen 5/9.

 

[etotheipi]

Yeah, it's a little odd. I wonder if they've just corrected for the moment of inertia of the physical ball bearings they've bought, in that maybe their density decreases a little bit with increasing radial distance, or something. I guess it's possible to check which coefficient matches the data best!

 

[kuruman]

I agree with the $\frac{5}{7}$ coefficient and the analysis for a ball that rolls without slipping. My question is, how does the lab manual present the equation for the acceleration. Is it a claim of theoretical proof (essentially what you provided) or a statement of fact as in "In the ramp that you will use, the linear acceleration is given by $\frac{5}{9}g\sin\theta.$" Steel balls are notorious for rolling with slipping especially on relatively steep ramps.

In any case, this is an experiment in which you have to measure the acceleration. Measure it and see what coefficient you get. No matter what we or the manual say, Nature wants it one way only. I am curious to know what that is so please let us know after you perform the experiment. There is no greater satisfaction than proving the manual wrong ... experimentally.

 

[berkeman]

I agree with the $\frac{5}{7}$ coefficient and the analysis for a ball that rolls without slipping. My question is, how does the lab manual present the equation for the acceleration. Is it a claim of theoretical proof (essentially what you provided) or a statement of fact as in "In the ramp that you will use, the linear acceleration is given by $\frac{5}{9}g\sin\theta.$" Steel balls are notorious for rolling with slipping especially on relatively steep ramps.

But if the ball slips a bit, that will increase the acceleration, not decrease it, no? It seems like the manual is saying that the acceleration is decreased for some reason from what is expected...

 

[jbriggs444]

I assume it is solud and uniform.
5/7 sounds very familiar and your analysis is correct. Never seen 5/9.

If one reverse-engineers the result, it suggests a moment of inertia of $\frac{4}{5}mr^2$. That's a mass distribution almost as far toward the rim of a thin hoop as one can get.

Or a factor of two error in a formula compared to the correct figure for a solid sphere of $\frac{2}{5}mr^2$.

 

[kuruman]

But if the ball slips a bit, that will increase the acceleration, not decrease it, no? It seems like the manual is saying that the acceleration is decreased for some reason from what is expected...

Yes, that is true. I thought of it too late to retract the post.

 

[etotheipi]

I agree with the 5/7 coefficient and the analysis for a ball that rolls without slipping. My question is, how does the lab manual present the equation for the acceleration

It does just say 'it can be shown that...' and offers no further explanation.

However, now having done the practical, it turns out it was slightly different to how it was presented on the diagram in the manual. Instead of a ramp with a flat surface, the ball-bearing rolled down an angle bar (i.e. a V-shaped track).

I couldn't measure the angle, but it's safe to say it was roughly a right angle. The ball-bearing is now in contact with the angle bar in two places, so there are two frictional forces. That doesn't change the moment of inertia of the ball-bearing, but it does change the torque vectors. Perhaps if the problem is worked through like this, we'd get their result?

I'm going to have tea now, but I'll try it later if no one else has beaten me to it!

 

[kuruman]

Ahh! An angled ramp will do it. You will need to do some trig and apply the parallel axis theorem to calculate the moment of inertia $I_P$ about the point of contact with the ramp. The acceleration will be less in this case because the lever arm of the torque is smaller. In the limit that the distance between the two points of contact is equal to the diameter, the acceleration goes to zero. Like I said, do the experiment and see what you get.

 

[etotheipi]

Yeah, I think it works! I aligned the $\hat{z}$ axis parallel to the track (pointing down the track), and the $\hat{x}$ and $\hat{y}$ vectors pointing vertically and horizontally in the plane of the cross-section.

Then, with the origin of coordinates at the centre of mass of the ball, I said$$\sum \tau_x = 2rF \cos{\theta} = I_{x} \alpha_x$$We also need to relate $\alpha_x$ to the acceleration $a_z$, which we can do by constraining zero relative velocity at the contact points,$$\vec{0} = \vec{v} + \vec{\omega} \times \vec{r} = \vec{v} + \alpha_x t \hat{x} \times (r\sin{(\theta)} \hat{x} - r\cos{(\theta)} \hat{y}) = \vec{v} - \alpha_x r t \cos{(\theta)} \hat{z}$$where $\theta$ is the angle of each edge of the "V" above the horizontal. Hence $a_z= \alpha_x r \cos{\theta}$, and$$mg\sin{\theta} - 2F = ma_z \implies mg\sin{\theta} = \frac{I_x a_z}{r^2\cos^2{\theta}} + ma_z = a_z \left( m + \frac{I_x}{r^2 \cos^2{\theta}} \right)$$or better,$$a_z = \frac{1}{\left(1+ \frac{I_x}{mr^2 \cos^2{\theta}} \right)} g\sin{\theta}$$If $\theta = \pi/4$, and $I_x = \frac{2}{5} mr^2$, then $\frac{I_x}{mr^2 \cos^2{\theta}} = \frac{4}{5}$ and $$a_z = \frac{5}{9} g\sin{\theta}$$

 

[jbriggs444]

That doesn't change the moment of inertia of the ball-bearing

It does, however, change the rotation rate of the ball bearing compared to distance moved!

If you consider the ball bearing to be a shape rolling down a track at $\frac{\sqrt{2}}{2}r$ from the center of the bearing, it is clear that some of the bearing's mass is further away from the center than this reduced r.

Indeed if we increase effective $r$ by a factor of $\sqrt{2}$ then this makes the moment of inertia come out as $\frac{4}{5}mr^2$, exactly as predicted.