Max Ratio of Particle Mass to Bowl Mass for No Slide

[Tonyt88]

A hemispherical bowl of mass M rests on a table. The inside surface of the bowl is frictionless, while the coefficient of friction between the bottom of the bowl and the table is μ = 1. A particle of mass m (and negligible size) is released from rest at the top of the bowl and slides down into it. What is the largest value of m/M for which the bowl will never slide on the table?

 

[Tonyt88]

And my teacher gives me the hint that the double angle formulas will be helpful, but I have no idea where those would be helpful b/c I would assume the only key moment is when the particle is at the base of the bowl.

 

[berkeman]

Wow, u=1? Are you sure that isn't a typo? Kind of a simple problem otherwise. Has to be a typo.

 

[Tonyt88]

Hmm, yes I didn't think u = 1 was very practical but I'm almost positive that it's not a typo, so then how it would it work?

 

[berkeman]

Sorry, I've never seen u=1. That means pinned, IMO.

 

[Tonyt88]

What do you mean pinned?

Well here's the link perhaps to clarify, it's #10:

http://isites.harvard.edu/fs/docs/icb.topic90598.files/hmwk3.pdf

 

[George Jones]

A hemispherical bowl of mass M rests on a table. The inside surface of the bowl is frictionless, while the coefficient of friction between the bottom of the bowl and the table is μ = 1. A particle of mass m (and negligible size) is released from rest at the top of the bowl and slides down into it. What is the largest value of m/M for which the bowl will never slide on the table?

Assuming the ball doesn't move, find the acceleration of the particle, which, since the speed of the particle is changing, is not strictly centripetal. There are two forces acting the particle - gravity and the contact force of the bowl on the particle. Use Newton's second law to find the contact force.

There are four forces acting on the bowl - gravity, static friction, the contact forcle of the particle on the bowl, and the contact force of the table on the bowl.

Sorry, I've never seen u=1. That means pinned, IMO.

No, it just means that in order to start the object sliding, the applied force must be greater than the normal force. For example, the coefficient of friction between the tires on drag racers and the track often is greater than one.

 

[Tonyt88]

I'm sorry, I'm still quite puzzled, is there anything else you can tell me about the problem that might aid me.

 

[George Jones]

What are the tangential and normal components of the particle's acceleration?

 

[Tonyt88]

tangential acc = d|v|/dt

normal acc = v^2/r = 2gh/r = Normal force of the bowl onto the ball ?

 

[George Jones]

tangential acc = d|v|/dt

normal acc = v^2/r = 2gh/r = Normal force of the bowl onto the ball ?

Yes.

Now, write the x and y component equations of F = ma. I would introduce an angle $\theta$ that is measured from horizontal, so, initially, $\theta = 0$ for the particle.