Determining the tension on a rotating particle

[freddie711]

Homework Statement

[/B]
A particle of mass m slides (both sideways and radially) on a smooth frictionless horizontal table. It is attached to a cord that is being pulled downwards at a prescribed constant speed v by a force T (T may be varying)

Use F=ma in polar coordinates to derive an expression for the tension T (T will depend on r and θ and how they may be changing)

Show the particle's polar coordinates satisfy r²dθ/dt = constant

HINT: The only horizontal force on the particle is T and it acts purely in the inward radial direction. Also, dr/dt is known and it is equal to -v

Homework Equations

a = (r''(t)w²)r_hat +(r(t)θ''(t) + 2r'(t)θ'(t))θ_hat
L(t) = L_o ?

The Attempt at a Solution

Not entirely sure how to get started. I've identified that a_r must be zero (since dr/dt is a constant, r''(t) must be zero making a_r zero). I've also set up the equation a_θ = r(t)θ''(t) - 2vw since r'(t) = v and θ'(t) = w. The issue I'm having is trying to identify another equation for the acceleration (or force) in the θ direction, and then from there rectifying that into T.

 

[TSny]

Welcome to PF!

If the force has only a radial component, what can you say about the component of acceleration in the $\small \theta$ direction: $a_{\theta}$?

What is the expression in polar coordinates for $a_{\theta}$?

 

[freddie711]

That's part that's confusing... if there's only a force radially, θ''(t) should be zero. However, intuitively, as the radius gets shorter the particle should start rotating faster (cons. of ang. mom.) As far as a_θ goes, I'm not exactly sure what the expression should be because I can't identify any forces working on the particle along that axis.

The other part that's messing with my head is the fact that, despite there being a force in the radial direction, there is no radial acceleration.

I've attached the image that came with the problem.

 

[TSny]

That's part that's confusing... if there's only a force radially, θ''(t) should be zero.

It turns out that $a_\theta$ is not equal to $\small \ddot{\theta}$. Note: From your first post you wrote a = (r''(t)w2)r_hat +(r(t)θ''(t) + 2r'(t)θ'(t))θ_hat. So, you can read off $a_\theta$ from there.

 

[freddie711]

Oh. That makes a lot of sense, thanks. How would I use that to get to T(t) though?

 

[TSny]

How would I use that to get to T(t) though?

Consider $a_r$. In your expression
a = (r''(t)w2)r_hat +(r(t)θ''(t) + 2r'(t)θ'(t))θ_hat
it doesn't appear that you have written the r-component correctly.
(Note: there is a tool bar for formatting. Click on the $\Sigma$ tab for math symbols.

You know that $r^2 \dot{\theta}$ is a constant, say $\small h$. So, $r^2 \dot{\theta} = h$.Try to find an expression for $a_r$ in terms of just $r$ and $h$. Then think about how $r$ changes with time.