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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.

- Thread starter NotMrX
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- #1

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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

berkeman

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- #3

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Assume no damping.

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dont forget the ball's moment of inertia

- #5

berkeman

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- #6

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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(1-cosZ) = (1/2)mv^2 + (1/2)(2/5)mR^2(v/R)^2

gR(1-cosZ) = 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(1-cosZ)

ma = m v^2/9R = mgsinz- F

F =mgsinz-mv^2/9R = mgsinz-(10/63)mg(1-cosZ)

Torque= I ddz/ddt = 9Rmg sinZ - 10Rmgsinz-(10/63)g(1-cosZ)

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

berkeman

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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

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Ei =-mg9Rcos(Zmax)berkeman said:

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.

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

berkeman

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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.

[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.

Last edited:

- #10

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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

berkeman

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- #12

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- #13

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Why do you not need to take into account the translational kinetic energy?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.

- #14

berkeman

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My post was from over 4 years ago. I don't remember that far back....Why do you not need to take into account the translational kinetic energy?

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