Spherical Shell with Fluid

In summary, A spherical shell filled with a frictionless fluid of equal mass is released from rest on an incline and rolls without slipping down the incline. The acceleration of the shell down the incline just after it is released is given by the equation a = 3/5*g*sin θ, where g is the acceleration of free fall and θ is the angle of the incline with the horizontal. This can be derived by considering the differences in density and inertia between a fluid filled sphere and a solid sphere. The fluid inside the sphere will rotate slightly relative to the room, but the faster the sphere moves, the less the fluid rotates due to the reduction of friction. Ultimately, the fluid will not rotate at all if it were
  • #1
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Homework Statement


A spherical shell of mass M and radius R is completely filled with a frictionless fluid, also of mass M. It is released
from rest, and then it rolls without slipping down an incline that makes an angle θ with the horizontal. What will
be the acceleration of the shell down the incline just after it is released? Assume the acceleration of free fall is g.
The moment of inertia of a thin shell of radius r and mass m about the center of mass is I =2/ 3mr2; the moment
of inertia of a solid sphere of radius r and mass m about the center of mass is I =2/5mr2.

(A) a = g sin θ
(B) a =3/4 g sin θ
(C) a =1/2 g sin θ
(D) a =3/8 g sin θ
(E) a =3/5 g sin θ.

Homework Equations

The Attempt at a Solution


I considered the fluid inside to be a "sphere in sphere" and used conservation of energy to get 2M*g*h=M*v^2 + 2/5*Mv^2+ 2/3 * Mv^2. (I used the fact that 2*1/2 =1 and that v=omega*r to simply). Adding the right hand side and diving by M give me 2*g*h= 29/15 v^2. I then divide by 2*h to get g=(29/15*v^2)/2h. How do I go from here?
 
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  • #2
postfan said:
2/5*Mv^2
How will a fluid filled sphere behave differently from a solid sphere of the same mass distribution? (Not a rhetorical question.)
 
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  • #3
Won't the density inside the the sphere be less than the exterior since there is the same mass with a far larger volume in the interior?
 
  • #4
postfan said:
Won't the density inside the the sphere be less than the exterior since there is the same mass with a far larger volume in the interior?
I wrote "same mass distribution'. I.e., what if the fluid were replaced by an equally dense solid, not the same as the sphere's shell, but fixed rigidly inside it?
 
  • #5
I would say then there would be no difference.
 
  • #6
postfan said:
I would say then there would be no difference.
No, there's a difference. Imagine a large basin filled with water in front of you. You seize the handles on the sides and turn it suddenly. Does it require a lot of torque?
 
  • #7
Yes?
 
  • #8
postfan said:
Yes?
Well, no it doesn't. Nowhere near as much as if it contained the same mass of ice. Can you not see why?
Fill a cup with water and sprinkle something on top that you can then observe floating. Rotate the cup. What do you notice? What would look different if the water were solid?
 
  • #9
I noticed that the floating thing is rotating as well and I think that if the water was solid that nothing would rotate.
 
  • #10
postfan said:
I noticed that the floating thing is rotating as well and I think that if the water was solid that nothing would rotate.
relative to the room or relative to the cup? (I mean, while the cup is rotating.)
 
  • #11
Relative to the cup.
 
  • #12
postfan said:
Relative to the cup.
Right, the liquid does not rotate fully with the cup. It does rotate a bit because of drag between the cup and water and within the water, but go back and read the problem statement. What does it tell you about the fluid here?
 
  • #13
It's frictionless.
 
  • #14
postfan said:
It's frictionless.
If the liquid in your cup was frictionless, would it rotate (relative to the room) when you rotated the cup?

What about the liquid inside the rotating sphere?
 
  • #15
Yes, so same with the sphere.
 
  • #16
It would rotate relative to the room if it were frictionless? If there's no friction between the cup and the liquid how does it get rotating?
 
  • #17
Well I thought that if it was frictionless that the it wouldn't rotate with respect to the cup and that since the cup is rotation with respect to the room that the liquid would also be in rotation relative to the room.
 
  • #18
postfan said:
Well I thought that if it was frictionless that the it wouldn't rotate with respect to the cup and that since the cup is rotation with respect to the room that the liquid would also be in rotation relative to the room.
By "it wouldn't rotate relative to the cup" you essentially mean "it would rotate with the cup" right?

The liquid has inertia, it takes some force (friction) to get it to rotate with the cup. Otherwise, it would rather sit still.
 
  • #19
Man all the reference frames can get confusing sometimes, but I'm starting to get what you're saying.
 
  • #20
postfan said:
Man all the reference frames can get confusing sometimes, but I'm starting to get what you're saying.
Consider a bowl of soup. If you rotate the bowl slowly, the soup rotates with it.
But what happens when you rotate the bowl quickly? The soup still rotates with the bowl slightly, but not nearly as much.
The faster you rotate the bowl, the less the soup rotates.

This is because rotating the bowl faster and faster is like reducing the force of friction. You're not actually reducing the force of friction, but you are reducing the time that the friction is acting for (thereby reducing the effect of friction). So you see, the more you reduce the friction, the less the soup rotates. If the soup was somehow frictionless, it would not rotate at all!

So back to the ball filled with liquid rolling down the hill (an interesting problem!).
It will be like a spherical shell rolling down a hill with a ball of liquid just "floating" down the hill (i.e. not rotating). See if you can't work out the mathematics of that situation.
 
  • #21
Okay so I came up with 2Mgh=.5Mv^2+/5*2/3*Mv^2+1/2mv^2 (once again using v=omega*r)
Adding and simplifying I got gh=2/3*v^2.
Where do i go from there?
 
  • #22
If you can explain your equations a bit that would be helpful (especially since it's hard to read without tex)

It looks like you used conservation of energy? Conservation of energy isn't necessary in this situation, just consider the forces on the object. What are all the forces/torques, and what effect should they produce?
 
  • #23
I did use conservation of energy.

But taking forces I see gravity acting down with a force of 2Mg and a normal force of 2Mgcos(theta). Am I missing something?
 
  • #24
postfan said:
Am I missing something?
What about friction? (Assume the coefficient of static friction is large enough.)

Remember the goal is to find the acceleration of the object down the slope.
 
  • #25
Wait, isn't friction negligible since it's a sphere?
 
  • #26
No. Friction is 100% necessary. Without friction, the sphere won't roll (this is because it is rolling down an incline).
 
  • #27
Why is that?
 
  • #28
That is what I would like you to figure out :)

What does it mean for the sphere to be "rolling," can you write it in an equation?
 
  • #29
What I'm thinking is that for a sphere to roll it need both a translational and angular velocity, but I'm don't know how to express that as an equation.
 
  • #30
postfan said:
What I'm thinking is that for a sphere to roll it need both a translational and angular velocity, but I'm don't know how to express that as an equation.
Right, but not just any translational/angular velocity.

To be "rolling" means there is a special relationship between the angular speed and the translational speed.

I will just tell you:

[itex]v=ωR[/itex]

What does this tell you about the translational and angular acceleration?
 
  • #31
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.
 
  • #32
postfan said:
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.
 
  • #33
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?
 
  • #34
postfan said:
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 [itex]F_f=\mu F_N[/itex] 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 Ff (or something) don't even bother with the normal force.

postfan said:
2Mgsin(alpha)-2Mgcos(alpha)tan(alpha)=ma

Rewrite it with an unknown force of friction Ff

postfan said:
ONow for torque:
2*M*gcos(alpha)*h/sin(alpha)=2/3MR^2*angularacceleration
It looks like you used conservation of energy :olduhh:

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

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 :sleep:
 
  • #35
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?
 

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