Advanced Rotational Velocity Question

[Victorzaroni]

Homework Statement

A particle of mass m rotates with a uniform speed on the inside of a bowl's parabolic frictionless surface in a horizontal circle of radius R=0.4 meters as shown below. At the position of the particle the surface makes an angle θ=20 degrees with the vertical. The angular velocity of the particle would be:

Homework Equations

?

The Attempt at a Solution

I have no idea what to do. The answer is 8.2rad/s

 

[PeterO]

Homework Statement

A particle of mass m rotates with a uniform speed on the inside of a bowl's parabolic frictionless surface in a horizontal circle of radius R=0.4 meters as shown below. At the position of the particle the surface makes an angle θ=20 degrees with the vertical. The angular velocity of the particle would be:

Homework Equations

?

The Attempt at a Solution

I have no idea what to do. The answer is 8.2rad/s

The object is traveling in a circle, so the net force is is horizontal towards the centre.
The acting forces - which combine to make that net force - are weight (vertically down) and the Normal Reaction force which acts at right angles to the slope, so is in a direction 20^o above the horizontal.

Draw that triangle of forces (as arrows as usual) and you will be able to use trig and/or Pythagoras to calculate the size of the centripetal acceleration.

From that you can get the angular velocity.

 

[Victorzaroni]

how would I express the normal reaction force? I've got Fc=Fnet, so ma_c=mg+N. What is N? Is it mgsinø, mgcosø? What is it?

 

[PeterO]

how would I express the normal reaction force? I've got Fc=Fnet, so ma_c=mg+N. What is N? Is it mgsinθ, mgcosθ? What is it?

I am not sure it is either. If the mass was sitting on a slope of that angle it would be one of them - but this mass is circulating, so the reaction force has to be much larger.

The weight force would be represented by an arrow, vertically down.
To add the Reaction Force, you draw its arrow starting at the end of the weight force arrow [join the arrows head to tail]
The Reaction arrow is angled up at 20 degrees.
It has to be long enough to make the resultant force horizontal.

Try drawing it - then analysing the resulting picture.