A ball rolls without slipping in a cylindrical trough

[Philethan]

Homework Statement

A solid sphere sphere (radius = R) rolls without slipping in a cylindrical trough (radius = 5R) as shown in the following figure. Show that, for small displacements from equilibrium perpendicular to the length of the trough, the sphere excuses simple harmonic motions with a period [tex]T=2\pi \sqrt{28R/5g}[/tex]

Homework Equations

[tex]\vec{F}_{net}=m\vec{a}[/tex]
[tex]\vec{\tau }_{net}=I\vec{\alpha }[/tex]
[tex]\left | \vec{a} \right |=R\left |\vec{\alpha }\right |[/tex]

The Attempt at a Solution

Step1. Draw the Free body diagram.

Step 2. Solve the Newton's Equation
[tex]Mgsin(\Omega )-fs=Ma[/tex]
Take the center of sphere as pivot.
[tex]f_{s}R=(\frac{2}{5}MR^{2})\alpha [/tex]
[tex]a=R\alpha [/tex]

Finally,
[tex]\therefore \, f_{s}=\frac{2}{7}Mgsin(\Omega )[/tex]

Step 3. Take the center of the cylindrical trough as the pivot.

[tex]\tau _{net}=[-Mg\times 4Rsin(\Omega )]+[\frac{2}{7}Mgsin(\Omega )\times 5R]=(\frac{2}{5}MR^{2}+M(4R)^{2})\frac{d^{2}\Omega }{dt^{2}}[/tex]
[tex]\therefore \: -\frac{18}{7}MgRsin(\Omega )=\frac{82}{5}MR^{2}\frac{d^{2}\Omega }{dt^{2}}[/tex]
[tex]\therefore \: -\frac{45}{287}\frac{g}{R}\Omega \approx \frac{d^{2}\Omega }{dt^{2}}[/tex]
[tex]\therefore \: T\approx 2\pi \sqrt{\frac{287R}{45g}}=2\pi \sqrt{\frac{31.\bar{8}R}{5g}}[/tex]

I've seen the correct solution but I still don't know why I was wrong.
I will be grateful for any help you can provide.

 

[thecommexokid]

When the ball is rolling down the trough, it is not pivoting about its center, nor about the center of the trough. It is, at every instant, pivoting about its point of contact with the surface.

 

[Nathanael]

When the ball is rolling down the trough, it is not pivoting about its center, nor about the center of the trough. It is, at every instant, pivoting about its point of contact with the surface.

I don't see how that's relevant. You can still treat the ball as moving translationally and rotationally about it's center of mass, which is a useful perspective to find the force of friction. The contact point between the ball and surface may be where it is purely rotating (that is, no translational motion) but it is valid to look at it at other perspectives as well.

 

[Quantum Defect]

I think that the key thing here is that there are two angles that are important. The angular displacement of the ball (what you call Omega) and the angular rotation of the ball (call it theta). In the first part you are talking about the angular acceleration of the ball, and in other parts you are talking about the angular accelaration of the ball on the ramp. I would try to come up with one equation, with a single angle: d^2(theta)/dt^2 + K theta = 0 [or with Omega] to get an equation for a SHO. The two angles are related in a simple way: R * Theta = 5*R*Omega

Working from your Eq. 2., using a single, consistent angle, I think that you should be able to come up with an equation...

 

[haruspex]

##\tau _{net}=[-Mg\times 4Rsin(\Omega )]+[\frac{2}{7}Mgsin(\Omega )\times 5R]=(\frac{2}{5}MR^{2}+M(4R)^{2})\frac{d^{2}\Omega }{dt^{2}}##

The right hand side is wrong. It treats the ball as rotating as a whole about the centre of curvature of the trough, and ignores the rotation involved in its rolling.

 

[Philethan]

The right hand side is wrong. It treats the ball as rotating as a whole about the centre of curvature of the trough, and ignores the rotation involved in its rolling.

Thanks a lot! At last, I understand why I was wrong! The net torque equation I wrote down just neglects the rotation involved in its rolling. However, that means I can't use the net torque equation about the center of curvature of the trough because that's meaningless. I can't use the net torque equation unless all points on the object has the same angular velocity about the pivot, right? Is this correct?

I guess I still can use the net torque equation because that's the derivative of kinetic energy with respect to θ.
Well, I still have some problem on it. Here's my reasoning.

Its total kinetic energy is
[tex]E_{k,total}=\sum_{i=1}^{n}\frac{1}{2}m_{i}v_{i}^{2}=\sum_{i=1}^{n}\frac{1}{2}m_{i}(v_{ic}+v_{c})^{2}=\frac{1}{2}Mv^{2}+\frac{1}{2}I(\frac{d\theta }{dt})^{2}[/tex]
where dθ/dt is the rotation rate of the ball about its center of mass. That is,
[tex]\frac{d\theta }{dt}\neq \frac{d\Omega }{dt}[/tex]
However, we still can find their relation through the instantaneous speed of center of mass of the ball:
[tex]v=4R(\frac{d\Omega }{dt})=R(\frac{d\theta }{dt})[/tex]

So, I can say its total kinetic energy is:
[tex]E_{k,total}=\frac{1}{2}M(4R\frac{d\Omega }{dt})^{2}+\frac{1}{2}(\frac{2}{5}MR^{2})(\mathbf{4}\frac{d\Omega }{dt})^{2}[/tex]
[tex]\therefore \: E_{k,total}=\frac{1}{2}[M(4R)^{2}+(\frac{2}{5}MR^{2})(\mathbf{4})^{2}](\frac{d\Omega }{dt})^{2}=\frac{1}{2}I_{eff}(\frac{d\Omega }{dt})^{2}[/tex]
[tex]\therefore \: \frac{dE_{k,total}}{dt}=I_{eff}(\frac{d\Omega }{dt})(\frac{\frac{d\Omega }{dt}}{d\Omega })=I_{eff}(\frac{d^{2}\Omega }{dt^{2}})=I_{eff}\alpha =\tau _{net}[/tex]

Now I can see why the right-hand side of my net torque equation was wrong, the true effective moment of inertia is
[tex]I_{eff}=M(4R)^{2}+(\frac{2}{5}MR^{2})(\mathbf{4})^{2}[/tex]
The factor 4 is the key point...

But when I plug it in the net torque equation, I found I'm still wrong.
[tex][-Mg\times 4Rsin(\Omega )]+[\frac{2}{7}Mgsin(\Omega )\times 5R]=[m(4R)^{2}+(\frac{2}{5}mR^{2})(\mathbf{4})^{2}](\frac{d^{2}\Omega }{dt^{2}})[/tex]
[tex]\therefore \: -\frac{18}{7}MgRsin(\Omega )=\frac{112}{5}MR^{2}(\frac{d^{2}\Omega }{dt^{2}})[/tex]
[tex]\therefore \: -\frac{45g}{392R}\Omega \approx (\frac{d^{2}\Omega }{dt^{2}})[/tex]
[tex]\therefore \: T\approx 2\pi \sqrt{\frac{392R}{45g}}=2\pi \sqrt{\frac{43.\bar{5}R}{5g}}[/tex]

That means my static frictional force is wrong, but why?...

Thanks a lot!edit:

If I calculate the "right static frictional force", then ..

[tex][-Mg\times 4Rsin(\Omega )]+[+f_{s}\times 5R]=(\frac{112}{5}MR^{2})\frac{d^{2}\Omega }{dt^{2}}[/tex]
[tex]\because \: T=2\pi \sqrt{\frac{28R}{5g}}\: \therefore \: -\frac{5g}{28R}\Omega \approx \frac{d^{2}\Omega }{dt^{2}}[/tex]
[tex]\therefore \: \frac{-20Mg\Omega +25f_{s}}{112MR}=-\frac{5g}{28R}\Omega[/tex]
[tex]\therefore \: f_{s}=0[/tex]

That is to say, the net torque about the center of mass of the ball is zero so that its acceleration should be zero. Why?...

 

[ehild]

It is difficult to follow your derivations. It would be better to use the CM frame of reference.
You also need to show what is the direction of motion of the centre of the ball, and what is the direction of its rotation about the centre.

The ball moves to the right and upward, and rotates about its centre clockwise with angular speed ω. fs has to decelerate rotation, so it points in the direction of the red arrow.
Equation for the motion of the CM:
## M 4R\frac {d^2\Omega}{dt^2}=-Mgsin \Omega+f_s##

Equation for the rotation about the CM
##f_sR=-I \frac{d\omega}{dt} ##

Rolling condition ##Rω=4R\frac{d\Omega}{dt} \rightarrow \frac{d\omega}{dt}=4\frac{d^2\Omega}{dt^2}##

and I=2/5 MR².

From the equation above you get ##\frac{d^2\Omega}{dt^2}=-\frac{5g}{28R}\Omega ## for Ω small, and fs is certainly not zero for Ω≠0.

 

[Quantum Defect]

Another approach, is to use an equation for (conservation of) energy:

1/2 M(v_com)^2 + 1/2 I omega^2 + Mg(h_com) = Mgh_max

use:
v_com = 4R dOmega/dt
omega = 4 dOmega/dt -- the 4 still puzzles me, seems like it should be 5R for the point of contact ??
h_com = 4R[1-cos (Omega) ]
I = 2/5 MR^2

Plug in the substitutions to get an expression solely in termos of Omega and dOmega/dt.

Take d/dt and use the fact that sin(x) = x for small x, and you will get an expression for simple harmonic motion with the period given above...

 

[Jediknight]

oh man, I think its time to get out the pencil and paper...this seems beyond me but only includes concepts I've covered-so in short its sure to be on the test

so you know your answer in the first post that your trying to show is right, right

I got to work through this before I help mush, but I know when gravity and shm are involved its usually because sin(θ)≈(θ) at small angles (I guess the last post already said that)

sorry if this seems off topic, a good practice question can be hard to come by so I just always like to give thanks

 

[haruspex]

But when I plug it in the net torque equation, I found I'm still wrong.
[tex][-Mg\times 4Rsin(\Omega )]+[\frac{2}{7}Mgsin(\Omega )\times 5R]=[m(4R)^{2}+(\frac{2}{5}mR^{2})(\mathbf{4})^{2}](\frac{d^{2}\Omega }{dt^{2}})[/tex]
[tex]\therefore \: -\frac{18}{7}MgRsin(\Omega )=\frac{112}{5}MR^{2}(\frac{d^{2}\Omega }{dt^{2}})[/tex]
[tex]\therefore \: -\frac{45g}{392R}\Omega \approx (\frac{d^{2}\Omega }{dt^{2}})[/tex]
[tex]\therefore \: T\approx 2\pi \sqrt{\frac{392R}{45g}}=2\pi \sqrt{\frac{43.\bar{5}R}{5g}}[/tex]

That means my static frictional force is wrong, but why?...

Thanks a lot!edit:

If I calculate the "right static frictional force", then ..

[tex][-Mg\times 4Rsin(\Omega )]+[+f_{s}\times 5R]=(\frac{112}{5}MR^{2})\frac{d^{2}\Omega }{dt^{2}}[/tex]
[tex]\because \: T=2\pi \sqrt{\frac{28R}{5g}}\: \therefore \: -\frac{5g}{28R}\Omega \approx \frac{d^{2}\Omega }{dt^{2}}[/tex]
[tex]\therefore \: \frac{-20Mg\Omega +25f_{s}}{112MR}=-\frac{5g}{28R}\Omega[/tex]
[tex]\therefore \: f_{s}=0[/tex]

That is to say, the net torque about the center of mass of the ball is zero so that its acceleration should be zero. Why?...

Sorry for the delay in replying. It took a while to figure out the flaw in your method.
In effct, you double-counted the frictional force. We need to understand what your calculated "effective moment of inertia" represents.
Forget gravity for the moment and imagine the ball as mounted on the end of a rod pivoted at the trough centre of curvature. Some agent applies a torque to the rod and observes the angular acceleration that results. This ratio is your effective moment of inertia. The agent is unconcerned with what is going on at the bottom of the rod.
So, when writing the left hand side of your equation, you should not subtract the torque of the frictional force from the applied torque. The effect of that force is already acounted for in the calculation of the effective moment of inertia.