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Ball in a bowl 
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#1
Aug306, 03:59 PM

P: 97

This problem has been driving me crazy. There is spherical ball of mass M and radius R. It is in a sphererical bowl with radius 10R. What is its frequency if pure rolling motion is assumed?
I figured it out if there is no friction so it slips. Sum of Torque = I ddz/ddt = Mg10RsinZ I = 1/2 MR^2 + M(9R)^2 W^2 = MG10R/(163/2MR^2) = (20 G)/(163 R) I am at a loss in figuring it out if the ball rolls. 


#2
Aug306, 04:16 PM

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P: 41,144

What force is causing the rolling motion? You need a force to move the ball and get some "frequency".



#3
Aug306, 04:22 PM

P: 97

Gravity pulls the ball down and we assume that it starts a small angle Z. We assume pure rolling so friction keeps it from slipping down the bowl.
Assume no damping. 


#4
Aug306, 06:18 PM

P: 156

Ball in a bowl
dont forget the ball's moment of inertia



#5
Aug306, 07:12 PM

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P: 41,144

Oh, so the ball starts at the waist of the inside of the sphere, and then rolls down and back up, kind of like a modified pendulum. Got it. As jasc15 points out, the main difference for this system versus a pendulum is the energy that goes into the rolling inertia of the ball. Write the equation for the energy of the ball as a function of height from the bottom of the arc, and include terms for the linear PE, linear KE and the energy in the rotation. That should start you on the way to working out the motion as a function of time, which gives you the natural frequency of the system.



#6
Aug406, 03:51 PM

P: 97

Ok. Here are mathmatical symbols for the system.
Ftangential = m a = mgsinZ  F(friction) Fcentripetal = ma = N mg cosZ Torque = I ddZ/ddt = 9Rmg sinZ  10R F(friction) E= mgrcosZ = (1/2)mv^2 + 1/2I w^2  mgR mgR(1cosZ) = (1/2)mv^2 + (1/2)(2/5)mR^2(v/R)^2 gR(1cosZ) = 7/10v^2 I really don't know what to do. This isn't for a class, I just study in my spare time. Taking the derivative twice doesn't seem like it would help. Further more the W in the equation refers to angular velocity of spinning and not the angular velocity of the ball going up and down. Maybe something along the lines of: v^2 = 10/7gR(1cosZ) ma = m v^2/9R = mgsinz F F =mgsinzmv^2/9R = mgsinz(10/63)mg(1cosZ) Torque= I ddz/ddt = 9Rmg sinZ  10Rmgsinz(10/63)g(1cosZ) w = (RMg/I)^(1/2) = (RMg/(2/5*MR^2 + M(9R)^2))^(1/2) w = (g/(R(2/5+81))^(1/2) Does this solution seem correct? Can we say a centipetal = v^2/R if it isn't a point mass? 


#7
Aug406, 04:52 PM

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P: 41,144

The first energy equation is almost correct, but I would write it more in the form of total energy TE = PE + KE(linear) + KE(rotational).
Then write an equation for the rotational velocity w of the ball as a function of position around its arc in the bowl  you have the TE and PE (just related to the vertical height, right?), so you can figure out the w from the TE equation. Once you have the w, that gives you the linear velocity v as a function of position around the bowl arc, and that will give you the position (angular displacement in the bowl) as a function of time. That gets you to the frequency number that you want to derive. 


#8
Aug706, 03:56 PM

P: 97

Ef = mg9R cos(Z) + 1/2mv^2 + 1/2 I w^2 = mg9R cos Z + 7/10Mv^2 v = [10/7*gR(cosZ  cosZmax)]^(1/2) I don't see how to make the velocity s function of time. The accelleration in not consant so kinematics won't work. 


#9
Aug706, 04:16 PM

Mentor
P: 41,144

I'm having trouble tracking your equations, so I'll try a couple in latex. I've sketched a semicircular bowl bottom of radius 10R, and I show the ball in two positions: (1) in the starting position at the top left of the bowl where the ball has only PE, call that position theta = +90 degrees, measured from the vertical, and (2) in the bottom position where the ball has all KE, and theta = 0 degrees.
[tex]PE(1) = 9MgR[/tex] [tex]KE(2) = \frac{1}{2} mv^2 + \frac{1}{2} Iw^2[/tex] [tex]KE(2) = \frac{1}{2} mv^2 + \frac{1}{2} * \frac{2}{5} mR^2 w^2[/tex] [tex]v = wR [/tex] [tex]KE(2) = \frac{1}{2} mv^2 + \frac{2}{10} mR^2 ({\frac{v}{R}})^2[/tex] Set PE(1) = KE(2) and solve for v at (2). That puts bounds on v(theta). Now write a more general equation for the KE as a function of theta (since you know what the PE is as a function of theta  just the height above the bottom of the bowl). EDIT  Latex makes that last term look like v^2/R, but it's obviously (v/R)^2. I'll try adding explicit parens there.... ahh, good. That worked. 


#10
Aug1006, 10:44 AM

P: 97

I am still trying!
Even for a simple pendulem it is only harmonic for small angles of Zmax. I just want to do simple harmonic motion; my differetial equations is limited. Part I: Energy at highest point: E =  mg(9R)cos(Zmax) Energy at lowest point: E = .5m(Vmax)^2 + .5Iw^2 mg9R = .5m(Vmax)^2 + .5(2/5)mR^2(v/R)^2 mg9R = (7/10)*m(Vmax)^2  mg9R Setting the energies equal: E=  mg(9R)cos(Zmax) = (7/10)*m(Vmax)^2  mg9R Solving energy equality for the velocity max: (Vmax)^2 = G9R*(1 cos(Zmax)) *(10/7) Part II: This is where I think I am making a mistake. S= (9R)*(Z) = (9R)*(Zmax sinwt) This starts it off at the bottum at time = 0. v = ds/dt = 9R*Zmax*w*coswt Vmax = 9R*Zmax*w Part III: combining the previous parts (Vmax)^2 = G9R*(1 cos(Zmax)) *(10/7) (9R*Zmax*w)^2 = G9R*(1 cos(Zmax)) *(10/7) Taylor series and small angles lead us to: cos(Zmax) = 1  (1/2)*(Zmax)^2 Substituting taylor series into eqn 9R*(Zmax)^2*w^2 = G[1  (1 (1/2)*(Zmax)^2)]*(10/7) = G*(Zmax)^2)*(10/7) Solving for w^2: w^2 =g/9R*(10/7) =(10 g)/(63R) The thing is i used a variation of a problem from the book and when i solved it with the same method I got a different answer. Does anyone see any fudamental mistakes? Also, what is Latex? I am on a public library computer and can't download programs. Sorry my math looks so mess. I wish i could draw pictures and write it neatly. 


#11
Aug1006, 11:28 AM

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P: 41,144

Why do you insist on calling the PE at the highest point negative? And why to you then subtract that energy out of the KE at the bottom of the bowl?



#12
Aug1006, 03:16 PM

P: 97

Thank you very much for your help. I got it to work out. The method was right I just made a algebra mistake. Anyways, I just use the center of the circle or bowl as height zero and that is why they are negative. The difference of the potential energy is the same where ever you call height zero.



#13
Feb211, 11:41 PM

P: 17




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