[thethinker]

Hi I have no idea how to do this question.

A hemispherical bowl, radius r, is resting in a fixed position with its rim horizontal. A particle P of mass m is moving in a horizontal circle around the smooth inside of the bowl. The centre of the circle is r/3 below the centre of the bowl. Fin d the angular speed of the particle and the magnitude of the reaction between the bowl and the particle.

[steve10]

Use trig to find the radius of the circle that the particle P is moving in.

Let the angular speed be w.

Express the centripetal force as a function of w.

Now the bowl is smooth.

The only force provided by the bowl will be a normal reaction at the point of contact.

Resolve this reaction into a (radially inwards) horizontal force and a vertical force.

Can you finish it off now?

[thethinker]

(Original post by steve10)
Use trig to find the radius of the circle that the particle P is moving in.

Let the angular speed be w.

Express the centripetal force as a function of w.

Now the bowl is smooth.

The only force provided by the bowl will be a normal reaction at the point of contact.

Resolve this reaction into a (radially inwards) horizontal force and a vertical force.

Can you finish it off now?

How do I use trig to find the radius of a circle? Thanks

The book doesn't have any example of this sort of question so I'm quite lost...

[browb003]

Should be like that

I remember doing this question, took me a while to realise that the horizontal radius that they've given in the picture goes all away round the hemisphere
I felt silly after realising ha

EDIT: I forgot to put r next to the number in the last 2 lines sorry

[thethinker]

(Original post by browb003)


Should be like that

I remember doing this question, took me a while to realise that the horizontal radius that they've given in the picture goes all away round the hemisphere
I felt silly after realising ha

EDIT: I forgot to put r next to the number in the last 2 lines sorry

Oops I didn't think about this haha thanks

[thethinker]

I'm unsure what to do now. Is F= mw^2 right

[steve10]

Nope, it's F = mw^2*r2, where r2 is the radius of the horizontal circle of motion. F is the centripetal force.

now call the reaction R, say, and resolve it into horizontal and vertical components, then do a balance of forces.

[thethinker]

(Original post by steve10)
Nope, it's F = mw^2*r2, where r2 is the radius of the horizontal circle of motion. F is the centripetal force.

now call the reaction R, say, and resolve it into horizontal and vertical components, then do a balance of forces.

So I've got Rcos x= mg and Rsinx= m*(2root2 over 3)r*w^2

Is that right? What next?

[steve10]

divide one by t'other

[thethinker]

(Original post by steve10)
divide one by t'other

So I get


tan x= (2rt2 r w^2)/g

What next? I've got the unknown of x?

[steve10]

almost there.

where did the 2root(2) over 3 come from ??

[thethinker]

(Original post by steve10)
almost there.

where did the 2root(2) over 3 come from ??

I don't know what you mean.

I tried this:

tanx= (r/3) / (2rt2 r/3)= 1/ 2rt2

so 1/ 2rt2 = 2rt2 r w^2/g

so g=rw^2

[steve10]

The method is right but could you check the below?

I get: Rsin x= mg and Rcosx= m*(2root2 over 3)r*w^2
(the sin and cos are switched about)

[thethinker]

(Original post by steve10)
The method is right but could you check the below?

I get: Rsin x= mg and Rcosx= m*(2root2 over 3)r*w^2
(the sin and cos are switched about)

Oh yeah that seems right

How do I do the last bit though?

Thanks for your continued help btw

[steve10]

The last bit - find R ?

Look at post #8 - use that

[thethinker]

(Original post by steve10)
The last bit - find R ?

Look at post #8 - use that

Cool I got w= rt(3g/r) which is in the answers

I don't know how to find R though..

[steve10]

From post #8 - Rcos x= mg
Actually, that should be - Rsin x= mg

Since you know that sin x = 1/3, you can solve for R.

Edit: put in wrong value for sin(x) - sorry!

[thethinker]

(Original post by steve10)
From post #8 - Rcos x= mg
Actually, that should be - Rsin x= mg

Since you know that sin x = 1/2rt(2), you can solve for R.

Actually sinx =1/3

But it worked! R=3mg

Thank you

[steve10]

Well done; Alles ist getan!