How Do You Calculate Velocity on a Circular Path Using Polar Coordinates?

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
4 replies · 2K views · last reply
nysnacc
Messages
184
Reaction score
3

Homework Statement


Diagram of a skier on a curved ski slope: a straight incline runs down to point A, then an arc of radius r sweeps through 45 degrees (0.25π rad) to point B, with the arc length labelled 6 m.
Homework problem screenshot, Part A: skier leaves a jump at B on a circular path of radius 6 m, asking for speed |v| with Value and Units input fields and a Submit button.


Homework Equations


s= r* theta
1/2 mv^2 = mgh energy conservation

The Attempt at a Solution


[/B]
I think I might be having something wrong in the attempt. And how do you do using the polar coordinate approach?

Handwritten work on lined paper: s = rθ gives 6 = r 0.25π, so r = 7.6394. Below, a circle of radius r with circumference 2rπ = 48 m, giving r = 48/2π = 7.6394, plus a small sketch of a circle with an arc labelled 6.
Handwritten energy conservation work on lined paper: 1/2 mv_A^2 + mgh_A = 1/2 mv_B^2 + mgh_B with h_B = 0, giving 17^2 + (9.81)(2.2375) = v_B^2, v_B = 17.633 m/s; a second attempt with v_g = sqrt(9.81 x 2.2375) = 4.6851 and a 45 degree circular arc sketch with r - h = r cos 45.
 
Physics news on Phys.org
Hey

The question gives r as 6m, not the arch length as 6m?

Cant help with much else tho! :)
 
OHHHH yes. r = 6m, so I can save time for getting that, good catch :)
And am I doing well on the energy conservation approach?
 
I can't see your working very clearly on this tablet I'm using but where does the 1/2 come from in 1/2*m*g*h
 
CWatters said:
I can't see your working very clearly on this tablet I'm using but where does the 1/2 come from in 1/2*m*g*h
I wrote it wrong, it shouldn't be 1/2 there