Particle moving inside a rotating hollow tube

[zeralda21]

Homework Statement

The hollow tube is pivoted about a horizontal axis through point O and is made to rotate in the vertical plane with a constant counterclockwise angular velocity θ'=3rad/s. If a 0.1-kg-particle is sliding in the tube toward O with a velocity of 1.2 m/s relative to the tube when the position θ=30 degrees is passed, calculate the magnitude N of the normal force exerted by the wall of the tube on the particle at that instant.

Homework Equations

Newtons second law.

The Attempt at a Solution

I decided to work with polar coordinates for this problem and therefore I have drawn two unit vectors, e_(r) and e_(θ), where e_(r) is along the tube(away from O) and e_(θ) is perpendicular to e_(r).

We know that θ'=3rad/s and that v=r'=-1.2m/s since the particle is moving towards O. It follows that r''=0 since r' is constant.

I have drawn a free-body diagram and for the particle there is:

A force mgcosθ in the negative e_(θ) direction, a normal force N(Isn't there a frictional force too?But I have neglected it since no information is given about it).

Since it doesn't move in the e_(θ)-direction it means that N=mgcosθ. Now I try to use it in Newton II:

F_(r)=ma_(r)=m(r''-r(θ'^2)) but I do not know how to proceed.

 

[mfb]

Can you draw a sketch of the system? At least for me, it is hard to imagine how the things are oriented and how the angle θ is defined.
Do we have gravity?

 

[zeralda21]

Can you draw a sketch of the system? At least for me, it is hard to imagine how the things are oriented and how the angle θ is defined.
Do we have gravity?

Yes, of course. I'll upload a picture. I'm sorry for confusing you.

Problem 3/53. http://i.imgur.com/N0Y7GCy.jpg

Source: Engineering Mechanics, Meriam Kraige

 

[mfb]

Thanks.

It follows that r''=0 since r' is constant.

This does not follow, r' is 1.2m/s at that specific point only.

If nothing is said about friction, you can neglect it.

Since it doesn't move in the e_(θ)-direction

It does move in that direction, as it has to stay in the tube.

 

[zeralda21]

Thanks.

This does not follow, r' is 1.2m/s at that specific point only.

If nothing is said about friction, you can neglect it.


It does move in that direction, as it has to stay in the tube.

Alright, some serious errors there. I will have to check it out thoroughly and get back.

 

[zeralda21]

Thanks.

This does not follow, r' is 1.2m/s at that specific point only.

If nothing is said about friction, you can neglect it.It does move in that direction, as it has to stay in the tube.

Alright, I see your point. I've tried this:

In the e_(θ) direction: N-mgcosθ=ma_(θ)=m(rθ''+2r'θ') where N is the normal force.

N = mgcosθ+m(rθ''+2r'θ')=m(gcosθ+(rθ''+2r'θ')).

We do not know what r is but we know that θ'=3rad/s. Does that imply that θ''=0? In that case we can proceed independent of r.

N=m(gcosθ+2r'θ')

 

[mfb]

θ''=0, right.

Looks good.

 

[Ahmed Al-shara]

θ''=0 because it was written : (with a constant counterclockwise angular velocity θ'=3rad/s ) that's why ... i think :)

 

[petera88]

Hello. I have this same problem and was able to find it here. I actually got everywhere up to the point of

N = m(2r'theta' +gcos(theta)) but am getting a wrong answer. I assumed this was the final answer. Since there is no normal force in the er direction or Ez direction, why isn't this it?

 

[Kameel]

Note that the r' is negative yielding negative acceleraion

 

[mfb]

The previous post was from 2014, I think it is best to let the thread rest.