How does liquid in a ball affect its rotational acceleration down a ramp?

Question

Suppose we have a shell with mass $M$ and radius $R$. If we let that roll without slipping down a ramp of angle theta to the horizontal, we can easily find the acceleration of the shell the instant the ball is let go.

Now, what if we completely fill the shell of mass $M$ and radius $R$ with a frictionless fluid of mass $M$ and let it roll down (without slipping) a ramp with angle theta? I have been thinking of this for a while but cannot seem to find what the instantaneous acceleration is as soon as the $\text{shell + liquid}$ is let go.

Answer 1

If your goal is to assume the fluid inside is friction-less, then consider a rotating hollow sphere with a non-rotating mass inside. Include the total mass of the shell and water in $m$, but only include the inertia from the shell in $I$. Secondly, if you want the acceleration then you can't rely on energy methods, and need to write a free body diagram in 2D. I've oriented the x-axis along the downward ramp direction, and the y-axis perpendicular to that.

• $a_x$ acceleration of the object's center along the downward ramp [$m/s^2$],
• $\alpha$ angular acceleration (about z-axis) of the object's center [$rad/s^2$],
• $f_x$ friction force between shell and ramp (along x-axis pointed opposite of $a_x$) [$N$],
• $m$ the total mass of the object (fluid+shell) [$kg$],
• $g$ acceleration due to gravity [$m/s^2$],
• $\theta$ angle between the ramp and the horizontal ground [$rad$],
• $I$ moment of inertia about the center of mass for the shell [$kg \cdot m^2$],
• $N=mg\text{cos}(\theta)$ normal force perpendicular to ramp surface (positive y-direction) [$N$]
• $R$ outer radius of the shell [$m$],
• $M_z$ moments about z-axis [$Nm$].

First solve for the friction force $f$ assuming that no slip occurs, implying the shell's angular acceleration is matched to the ball's center acceleration, $$a_x=\alpha R$$ Summing the moments (about z-axis) to solve for friction force $f$ $$\sum M_z = I\alpha = fR$$

$$f = \frac{Ia_x}{R^2}$$

For completeness here is the y-direction equation of motion, though it's not needed: $$\sum F_y= 0 = N - mg \cdot \text{cos}(\theta)$$ Next, create the x-direction equation of motion: $$\sum F_x = ma_x = mg \cdot \text{sin}(\theta)-f$$ Substitute in the previous $f$ and solve for $a_x$:

$$ a_x = \frac{mg \cdot \text{sin}\theta}{m+I/R^2}$$

Answer 2

One way to approach this type of problem is to use conservation of energy

Let:

• $m=2M$ be the total mass of the object (fluid+shell)
• $g$ be the acceleration due to gravity,
• $h$ be the vertical distance the object moved from start,
• $v$ be the speed of the object down the ramp, and
• $I$ be the moment of inertia about the center of mass for the shell.
• $R$ be the radius of the shell.

I assume that the shell rolls without slipping, and that the fluid does not interact with the rolling shell. The latter is artificial, and probably could be alleviated, to first order, by increasing the effective moment of inertia of the shell and/or including a frictional force. Leaving these aside, we get

$mgh=\frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 $

$\omega = \frac{v}{R}$

Answer 3

If the fluid inside the shell is frictionless and without bubbles or sloshing, it may be treated as a solid which does not rotate, but just slides down the ramp. This provides one component of instantaneous acceleration, which you would add to the instantaneous acceleration of the hollow ball. It seems to me that with this method, you needn't combine the mass of the shell and the water into one computation. You can combine the results of two simpler computations.