What is the Acceleration of a Spherical Shell Filled with Fluid on an Incline?

[postfan]

Oh that's the equation, I've been using it the whole time without full understanding it. So that means then the angular acceleration = the translational acceleration divided by the radius.

 

[Nathanael]

So that means then the angular acceleration = the translational acceleration divided by the radius.

Right. So write an equation for the translational acceleration of the object involving gravity and friction, and then write an equation for the angular acceleration involving the torque on the sphere, and then use that relationship you just told me to put it all together.

 

[postfan]

Ok since I wasn't given a mu I just let mu=tan(alpha)

So balancing forces in the x direction:
2Mgsin(alpha)-2Mgcos(alpha)tan(alpha)=ma
However this implies that the acceleration is zero which seems odd unless the acceleration is purely angular.

Now for torque:
2*M*gcos(alpha)*h/sin(alpha)=2/3MR^2*angularacceleration

Can you tell me if my equations are right or not?

 

[Nathanael]

Ok since I wasn't given a mu I just let mu=tan(alpha)

You don't need (mu) the coefficient of friction. When rolling, the contact point of the sphere is stationary (the instantaneous velocity of the contact point is zero) therefore the friction is static friction. When the force of friction is static, the equation F_f=\mu F_N only represents the maximum force of friction possible. The actual force of friction could be anything below that. (That is why I said in my other post, assume the coefficient of static friction is "large enough").

So whenever you need the force of friction, just call it F_f (or something) don't even bother with the normal force.

2Mgsin(alpha)-2Mgcos(alpha)tan(alpha)=ma

Rewrite it with an unknown force of friction F_f

ONow for torque:
2*M*gcos(alpha)*h/sin(alpha)=2/3MR^2*angularacceleration

It looks like you used conservation of energy

The force of friction provides a torque on the ball right? You can write the torque using the unknown force of friction F_f

Then you will need to divide the torque by the rotational inertia (but be careful! remember page 1 of this thread) and that will give you the angular acceleration.

Then you can finally use the equation "angular acceleration = translational acceleration divided by R"I should go to sleep now

 

[postfan]

Ok so 2*M*g*sin(alpha)-F_f=2M*a for forces
and F_f*R=(2/3)*M*R^2*angularacceleration for torque

How's that?

 

[postfan]

Is that right ?

 

[Nathanael]

Ok so 2*M*g*sin(alpha)-F_f=2M*a for forces
and F_f*R=(2/3)*M*R^2*angularacceleration for torque

How's that?

It looks right. Then solve the second equation for the angular acceleration and multiply both sides by R and then plug that expression in for a (because Rα=a like we said) then you can solve for F_f and then you can put F_f back into the original linear acceleration equation to get the answer.

About your earlier question of why friction is necessary: If you placed the ball from rest on a frictionless ramp it would just slide down the ramp without rotating (there would be no torque).

 

[postfan]

So I did that and got a=3/4gsin(alpha). Is that right?

 

[Nathanael]

So I did that and got a=3/4gsin(alpha). Is that right?

Yes

 

[postfan]

Cool thanks for the help and staying up with me last night (till 4:30!).

 

[postfan]

Also did they just give you the moment of inertia for a solid sphere just to trick you?

 

[Nathanael]

Also did they just give you the moment of inertia for a solid sphere just to trick you?

Well, I can't get inside the mind of the test-makers, but if you're asking if that piece of information was irrelevant, then yes, it was irrelevant.

 

[haruspex]

did they just give you the moment of inertia for a solid sphere just to trick you?

Had they given you all the formulae except that one it would have been rather a strong clue.