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I am working on an experiment to learn about rotation motion and find the correlation behind the between the spheres’ diameters and the time taken for them to make their way down an track at constant length, L, held at a height h. Different diameters, d, of spheres were used. They sit on the rail with inner width, w.

I've use the laws of energy conservation. So the gravitational potential energy will be transferred mechanically to rotational and translational kinetic energy.

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

Since the ball rests on two points on the rail, the effective radius, R, of the sphere was worked out using the Pythagorean Theorem.

$$R^2 = r^2-\left(\frac{w}{2}\right)^2$$

And the moment of inertia of a solid sphere is $I = \frac{2}{5}mR^2$.

Substituting the equation for the moment of inertia into the equation for the conservation of energy. I got the following expression:

$$gh=v^2\left(\frac{1}{2}+\frac{1}{5}\frac{R^2}{r^2}\right)$$

Because I measured the time taken for the ball to travel the distance, L and v, in the equation above, refers to the final linear velocity of the sphere. I approximately the final linear velocity by the following equation.

$$\overline{v}=\frac{v}{2}$$

Finally, I substituted the expression for R and and v to create an expression in terms of r and t respectively.

$$\left(2\frac{L}{t}\right)^2=\frac{gh}{0.5+0.2\frac{r^2-(\frac{w}{2})^2}{r^2}}$$

Rearranging for t, I got the expression, $t=\frac{L(11.2r^2-0.8w^2)^{1/2}}{2(gh)^{1/2}r}$. This yields the graph below when plotted on desmos, it looks wildly different from the shape of the graph when I collected my raw data.

Desmos plot

Please let me know if there has been something wrong! Any advice on how to linearise the data using a log-log transformation would also be appreciated.

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2 Answers 2

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You should have defined the r and R in your picture the other way around. Humans prefer for $r<R$, and you have also misused some formulæ that assumed that too. In particular, $I=\frac25mR^2$ assumes the opposite of your convention, and I will also use the convention where it works.

I have no idea where you plucked out $\bar v=v/2$. You probably should be using the no-slip condition. After all, that is the condition that conserves energy in the rolling case. $v=r\omega$ $$mgh=\frac12mv^2+\frac12I\omega^2=\frac12mv^2+\frac15mR^2\left(\frac vr\right)^2\\ v = \sqrt{\frac{10gh}{5+2R^2/r^2}}$$ At this point, if you are only considering the rolling down this one inclined plane, you might use the $\bar v=v/2$

Do you think you need to take into account that dropping at different angles, i.e. different $h$ will change the stopping conditions by a little bit?

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  • $\begingroup$ Thank you! So I am to take R as the actual radius and r as the effective radius? I’m just a bit confused at why the expression for rotational inertia goes against the convention. Also the angles are kept constant in the experiment, so will I have to use the coefficient of static friction to set a no-slip condition? $\endgroup$ Apr 27, 2023 at 10:24
  • $\begingroup$ Yes, R is the actual radius and r the effective radius. I think the convention that is common is the one I am describing. Yes, static friction causes no-slip, but it tends to not appear in the equations because even a tiny amount of static friction tends to suffice. $\endgroup$ Apr 28, 2023 at 2:26
  • $\begingroup$ I believe I understand it now. I managed to obtain an expression of t in terms of the ratio w/R, with the limits w/R -> 0 and w/R -> 2. I think that’s consistent with the literature I have read and the no slip condition as the sphere will no longer be resting on the track (the width of the track is equal to the diameter of the sphere)? I’m just struggling to obtain a final expression with R isolated which can be matched to my experimental graph. Any suggestions? $\endgroup$ Apr 28, 2023 at 6:04
  • $\begingroup$ What is your expression for T in terms of R? $\endgroup$ Apr 28, 2023 at 6:06
  • $\begingroup$ I have gotten the equation expression for T in terms of R $\endgroup$ Apr 28, 2023 at 7:08
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you can obtain the time that take the sphere to roll down on the inclined track with these equations

$$ m\,a=m\,g\,\sin(\phi)+F_l+F_r\tag 1$$ $$I_s\,\dot\omega=R\,(F_l+F_r)\tag 2$$ the rolling condition $$a=R\,\dot\omega\tag 3$$ plus the condition that the sphere don't rotate about the z-axes

$$\frac w2\,(F_l-F_r)=0\tag 4$$

we obtain 4 equations for the unknows $~a,~\dot\omega~,F_l~,F_r~$

the solution for a is: $$a:=\frac{d^2 s(t)}{dt^2}=\frac{R^2}{m\,R^2+I_s}\,m\,g\,\sin(\phi)$$

the solution of this differential equation with $~s(0)=0~,\dot s (0)=0~$ is

$$s(t)=\frac 12 \frac{R^2}{m\,R^2+I_s}\,m\,g\,\sin(\phi)\,t^2$$

thus the time $~t^2~$ with $~s(t)=L~$ is

$$t^2=\frac 25\,{\frac {L \left( 7\,{r}^{2}-5\,{w}^{2} \right) }{g\sin \left( \phi \right) {R}^{2}}}\quad\text{with}\\ I_s=\frac 25\, m\,r^2\quad,R^2=r^2-w^2 $$

  • $\phi~$ incline angle
  • $F_l~,F_r~$ left and right constraint forces enter image description here

Plot $$t^2=\frac{14\,L}{5\,g\sin(\phi)}\frac{x^2-5/7}{x^2-1}\quad,x=\left(\frac rw\right)$$ enter image description here

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  • $\begingroup$ Thank you so much for your help! But I'm afraid that differential equations are beyond my level of understanding at the moment. However, I would love to look into the topic. I have an understanding of your explanation just before the solving of the differential equation. Do you have any resources on that part so I can have a better idea of how to go about it? $\endgroup$ Apr 28, 2023 at 13:46
  • $\begingroup$ Forgot the differential equation. The acceleration a is $ dv/dt $ thus to obtain the velocity you integrate the acceleration, analog to obtain the distance s you integrate the velocity $\endgroup$
    – Eli
    Apr 28, 2023 at 14:12
  • $\begingroup$ I see! Can I ask what kind of relationship would the expression be called, specifically where t^2 seems to be proportional to (x^2-5/7)/(x^2-1)? $\endgroup$ Apr 28, 2023 at 14:32
  • $\begingroup$ substitute $~x^2\mapsto y~$ this mean that the time t is proportional to square root of y $\endgroup$
    – Eli
    Apr 28, 2023 at 14:45
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    $\begingroup$ Thanks, I think understand it better now! $\endgroup$ Apr 28, 2023 at 15:01

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