How to Calculate Work Done by a Motor Pulling a Cup through a Parabola?

[Melawrghk]

Homework Statement

This isn't actually a homework question (well it is for homework). Basically for one of our projects we had a motor pull a cup through an upside down parabola shape. The motor was located off-axis in comparison to the parabola. The cup started from zero, then as the motor rotated it would wind the string and thus pull the cup.
Now I need to calculate the work done on the cup by the motor (neglecting friction and such).

The Attempt at a Solution

I'm not sure what to do here.
I thought of getting equations for both accelerations (tangential and normal), then combining them into just one acceleration. integrating over all radii (it decreases) to get the total force exerted, but I'm not sure it's additive? Then maybe multiply that total force by the arc length of the parabola?

Either way, I'm just looking for suggestions on how to do this.

 

[tiny-tim]

Basically for one of our projects we had a motor pull a cup through an upside down parabola shape. The motor was located off-axis in comparison to the parabola. The cup started from zero, then as the motor rotated it would wind the string and thus pull the cup.
Now I need to calculate the work done on the cup by the motor (neglecting friction and such).

Hi Melawrghk!

Work done = force dot distance,

so what are the forces on the cup, and what are the "dots" with distance?

 

[Melawrghk]

Hey thanks for the reply!

Well for forces, there are weight, tension in the string, normal force (from the rail that makes up the parabola). I only know the weight though...

As for the dots in distance, there are i and j. 'i' would be the horizontal span of the parabola (it's like 20 cm, I think) and 'j' would be zero? Because it returns to the same height as it started. Or is it the absolute value (in which case it's double the height)?

 

[tiny-tim]

Well for forces, there are weight, tension in the string, normal force ...

As for the dots in distance, there are i and j. 'i' would be the horizontal span of the parabola (it's like 20 cm, I think) and 'j' would be zero? Because it returns to the same height as it started. Or is it the absolute value (in which case it's double the height)?

Slow down!

To find the work done , it's easier to "dot" each force separately (especially since the direction of two of the forces is changing).

For the weight, I would choose i and j as you suggest.

But for the normal force and the tension, choose i and j to be the tangent and normal directions.

 

[Melawrghk]

Hah, okay :D

So for weight, it's just 'j'.
And for tension and normal I'd tilt (rotate, I suppose) the axis.

Since tension and normal are changing, does that mean I'll need an equation for each, relating it to the distance from the point of rotation? (or maybe the x-coordinate, now that I think of it)
Could I then find work done by each force when the cup is moved over an infinitesimal piece of parabola and integrate over the whole thing?

Thanks so much for the help!

 

[tiny-tim]

Since tension and normal are changing, does that mean I'll need an equation for each, relating it to the distance from the point of rotation?

what rotation? this isn't torque

Could I then find work done by each force when the cup is moved over an infinitesimal piece of parabola and integrate over the whole thing?

Yes, you find the infintesimal work done at each point, and then integrate.

 

[Melawrghk]

Sorry, bad wording... I meant, as the cup moves through the parabola, the radius will change. I don't really remember why I called it rotation...

Awesome! Thanks, I'll try that :)