M3 Motion in a circle

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?

Reply 2

OP

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 :smile:

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

Reply 3

bencwbrown

11

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 :tongue:

I felt silly after realising ha

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

(edited 11 years ago)

Reply 4

OP

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 :tongue:

I felt silly after realising ha

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

Oops I didn't think about this :facepalm2:

haha thanks :biggrin:

Reply 5

OP

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

Reply 6

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.

(edited 11 years ago)

Reply 7

OP

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?

Reply 8

divide one by t'other

Reply 9

OP

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?

Reply 10

almost there.

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

Reply 11

OP

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

Reply 12

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)

Reply 13

OP

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 :smile:

How do I do the last bit though?

Thanks for your continued help btw

Reply 14

The last bit - find R ?

Look at post #8 - use that

Reply 15

OP

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 :smile:

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

Reply 16