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blalien
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[SOLVED] Classical mechanics: ball rolling in a hollow sphere
This problem is from Gregory:
A uniform ball of radius a and centre G can roll without slipping on the inside surface of a fixed hollow sphere of (inner) radius b and centre O. The ball undergoes planar motion in a vertical plane through O. Find the energy conservation equation for the ball in terms of the variable [tex]\theta[/tex], the angle between the line OG and the downward vertical. Deduce the period of small oscillations of the ball about the equilibrium position.
So in summary, we have:
a: radius of ball
m: mass of ball
[tex]\theta[/tex]: angle of the ball's position, relative to the vertical line connecting the center and bottom of the hollow sphere
I: moment of inertia of ball
[tex]\omega[/tex]: rotational velocity of ball
T: kinetic energy of ball
V: potential energy of ball (V=0 at height [tex]\theta[/tex]=[tex]\pi[/tex]/2, the center of the sphere)
E: total energy of ball
g: acceleration due to gravity
I = 2/5ma^2
First of all, I'm assuming that [tex]\omega[/tex]=[tex]\theta[/tex]'. It sounds intuitive, but I could be wrong there.
I'm given, as a solution, that the period of small oscillation (that is, sin([tex]\theta[/tex])=[tex]\theta[/tex]) is 2[tex]\pi[/tex](7(b-a)/5g)^(1/2), which I'm not getting in my results. I have a very strong hunch that my mistake comes from bad energy equations. So, would you mind taking a look of these?
T = 1/2mv^2 + 1/2I[tex]\omega[/tex]^2
v = [tex]\omega[/tex]*(b-a)
So T = 1/2m([tex]\omega[/tex]*(b-a))^2 + 1/2(2/5ma^2)[tex]\omega[/tex]^2
T = m[tex]\omega[/tex]^2/10(7a^2-10ab+5b^2)
V = -(b-a)mgcos([tex]\theta[/tex])
So E = T + V = that stuff
Am I correct here?
Homework Statement
This problem is from Gregory:
A uniform ball of radius a and centre G can roll without slipping on the inside surface of a fixed hollow sphere of (inner) radius b and centre O. The ball undergoes planar motion in a vertical plane through O. Find the energy conservation equation for the ball in terms of the variable [tex]\theta[/tex], the angle between the line OG and the downward vertical. Deduce the period of small oscillations of the ball about the equilibrium position.
So in summary, we have:
a: radius of ball
m: mass of ball
[tex]\theta[/tex]: angle of the ball's position, relative to the vertical line connecting the center and bottom of the hollow sphere
I: moment of inertia of ball
[tex]\omega[/tex]: rotational velocity of ball
T: kinetic energy of ball
V: potential energy of ball (V=0 at height [tex]\theta[/tex]=[tex]\pi[/tex]/2, the center of the sphere)
E: total energy of ball
g: acceleration due to gravity
Homework Equations
I = 2/5ma^2
The Attempt at a Solution
First of all, I'm assuming that [tex]\omega[/tex]=[tex]\theta[/tex]'. It sounds intuitive, but I could be wrong there.
I'm given, as a solution, that the period of small oscillation (that is, sin([tex]\theta[/tex])=[tex]\theta[/tex]) is 2[tex]\pi[/tex](7(b-a)/5g)^(1/2), which I'm not getting in my results. I have a very strong hunch that my mistake comes from bad energy equations. So, would you mind taking a look of these?
T = 1/2mv^2 + 1/2I[tex]\omega[/tex]^2
v = [tex]\omega[/tex]*(b-a)
So T = 1/2m([tex]\omega[/tex]*(b-a))^2 + 1/2(2/5ma^2)[tex]\omega[/tex]^2
T = m[tex]\omega[/tex]^2/10(7a^2-10ab+5b^2)
V = -(b-a)mgcos([tex]\theta[/tex])
So E = T + V = that stuff
Am I correct here?