Circular Motion - particle in a bowl

1. Sep 1, 2008

Thanuka

For the above problem i got the answer. But i got another question

I really apprecaite if any one can give an answer for this question
This is the question

A particle of mass 0.3kg moves with an angular velocity of 10rad per second in a horizontal circle 20cm inside a smooth hemispherical bowl. FInd the reaction of the bowl on the particle and the radius of the bowl.

Here i found the centripetal of acting on the object
F=0.3*10^2*20/100
F=6N

Is this the reaction?
After this i don't know how to find the radius

2. Sep 1, 2008

Staff: Mentor

You found the centripetal force acting on the particle. That's only one component of the reaction force.

Draw a diagram of the particle showing all the forces acting on it. Hint: What can you deduce from the fact that the bowl is smooth?

3. Sep 1, 2008

Thanuka

ohhh yes if it is smooth centripetal force is the only componant of the reaction force as there is no friction
But how can i find the radius of the bowl??

4. Sep 1, 2008

Staff: Mentor

True, there is no friction, but that does not mean that the reaction force is horizontal. Hint: Consider the vertical forces acting on the particle.
First find the direction of the reaction force, then use some trig.

5. Sep 1, 2008

Thanuka

I think the centripetal force we found should be normal to the inner surface of the bowl
as centripetal acceleration always goes through the center of the circle ,,,i got a triangle which one side is 20cm and its hypotenus is the centripetal force which is 6N
After that i got stucked..

6. Sep 1, 2008

Thanuka

ohh sorry its not the centripetal acceleration its the centriprtal force that goes through the center of the bowl

7. Sep 1, 2008

Staff: Mentor

But the circle is horizontal, thus the centripetal force and acceleration will be horizontal.

The force that the bowl exerts on the particle is normal to the surface of the bowl. What component of that force equals the centripetal force?

8. Sep 1, 2008

Thanuka

I think we have to resolve the normal reaction force excert by the bowl on the particle like "Rcos theta"
then it is equal to the 6N which is centripetal force
I don't know ,,,is it right??

9. Sep 1, 2008

Staff: Mentor

Good. Now analyze the vertical components.

10. Sep 1, 2008

Thanuka

i think i got now
Rcos theta= 6N as centripetal force is 6N
Rsin theta = 3N as mass of the particle is 0.3kg
therefore:
tan theta = 1/2
theta = 26 deg and 34 mintues
then
from the triangle
we know theta therefore we can find cos theta
cos theta = 0.8954
By solving this you can get the lenght of radius
I hope this is right
Is it?????

11. Sep 1, 2008

Staff: Mentor

I didn't check your calculations, but that's exactly the right idea. Good!

12. Sep 1, 2008