Angular acceleration at the bottom

[kolua]

Homework Statement

You hold a small ice cube near the top edge of a hemispherical bowl of radius 100 mm. You release the cube from rest. What is the magnitude of its acceleration at the instant it reaches the bottom of the bowl? Ignore friction.

Homework Equations

a=v^2/r delta U= delta Krot +delta Ktrans

The Attempt at a Solution

mgR=1/2Iw^2+1/2mv^2
a=v^2/rSo I'm not sure if it is so far so good. How do I find the rotational inertia here? How should I approach this question.

 

[andrewkirk]

You can infer from the question saying 'small ice cube' that they want you to ignore any rotational energy of the cube. Such energy will be trivial if it is small, and cannot be calculated without knowing the size of the cube.

 

[Daeho Ro]

It's not usual rotation problem, because of the source of force is only gravity. As you know, it always acts toward ground, so you don't have to consider the rotational things.

 

[kolua]

Homework Statement

You hold a small ice cube near the top edge of a hemispherical bowl of radius 100 mm. You release the cube from rest. What is the magnitude of its acceleration at the instant it reaches the bottom of the bowl? Ignore friction.

Homework Equations

a=v^2/r delta U= delta Krot +delta Ktrans

The Attempt at a Solution

mgR=1/2Iw^2+1/2mv^2
a=v^2/rSo I'm not sure if it is so far so good. How do I find the rotational inertia here? How should I approach this question.

does a=2g sound good?

 

[Daeho Ro]

There is no horizontal forces acting the ice cube and only exist the vertical forces. The source of vertical force is only one. That is gravity.

 

[ehild]

does a=2g sound good?

Yes, it is correct.

 

[kolua]

Yes, it is correct.

Daeho Ro says it's only g?

 

[ehild]

You do not need to believe anybody. How did you get your result?

 

[ehild]

There is no horizontal forces acting the ice cube and only exist the vertical forces. The source of vertical force is only one. That is gravity.

There is not only gravity acting on the ice cube in the bowl. What do you think it is, and in what direction, up or down?

 

[kolua]

There is not only gravity acting on the ice cube in the bowl. What do you think it is, and in what direction, up or down?

I think there's also a centripetal force that's upward. And that give the Fnet=mg=F-mg F=2mg=ma a=2g?

 

[ehild]

I think there's also a centripetal force that's upward. And that give the Fnet=mg=F-mg F=2mg=ma a=2g?

The centripetal force does not "act": it is the resultant force needed to maintain circular motion with given radius and velocity. The wall of the bowl and the ice cube are in contact. Do they not exert force to each other? How this force is called?

 

[kolua]

The centripetal force does not "act": it is the resultant force needed to maintain circular motion with given radius and velocity. The wall of the bowl and the ice cube are in contact. Do they not exert force to each other? How this force is called?

Yes. I think's it's the contact force?

 

[ehild]

Yes. I think's it's the contact force?

Yes, or normal force. What do you think its magnitude is now?

 

[Daeho Ro]

The source of vertical force is only one. That is gravity.

I think my explanation is somewhat odd. Ignore this thing.

The forces acting on the ice cube are gravity and normal force. Of course, the normal force is larger than the gravity and the difference of these forces is acting as the centripetal force.

 

[kolua]

Yes, or normal force. What do you think its magnitude is now?

a=2g=19.6m/s^2?

 

[ehild]

You gave the acceleration. I asked the normal force.

 

[kolua]

You gave the acceleration. I asked the normal force.

3mg

 

[ehild]

Yes. Upward or downward?

 

[kolua]

Yes. Upward or downward?

upward