Riding a bicycle around a tree w/ constant acceleration

[Fernando Calvario]

Homework Statement

African baobab trees can have circumferences of up to 43.0 m. Imagine riding a bicycle around a
tree this size. If, starting from rest, you travel a distance of 162 m around the tree with a constant
angular acceleration of 5.00 × 10–2 rad/s2, what will your final angular speed be?

Homework Equations

circumference = 2*pi*r ; ω = v/r

The Attempt at a Solution

I thought of isolating r from the circumference and I got 6.844m. I planned using this for angular speed equation but then I had to deal with angular acceleration. At which point, I got confused given how I am a novice at this

 

[haruspex]

had to deal with angular acceleration

Angular motion has close parallels with linear motion. What equation would you use if you were given a distance, a standing start, and a linear acceleration and wanted to find the final velocity?

 

[Fernando Calvario]

Angular motion has close parallels with linear motion. What equation would you use if you were given a distance, a standing start, and a linear acceleration and wanted to find the final velocity?

vf ^2 = vi^2 + 2aΔx

 

[haruspex]

vf ^2 = vi^2 + 2aΔx

Right. Try converting that to angular entities.

 

[Fernando Calvario]

So I did this: ωf = √2(5*10^-2 rad/s^2)(162m-43m). I got 3.446 m/s

 

[haruspex]

So I did this: ωf = √2(5*10^-2 rad/s^2)(162m-43m). I got 3.446 m/s

Not so fast... let's get the general equation right first. Write it all in terms of angular quantities, i.e. no linear velocities, linear accelerations or linear displacements.

 

[Fernando Calvario]

Not so fast... let's get the general equation right first. Write it all in terms of angular quantities, i.e. no linear velocities, linear accelerations or linear displacements.

ωf² = ωi² + 2a (θf - θi)

 

[haruspex]

ωf² = ωi² + 2a (θf - θi)

What is a there? Angular acceleration is usually written α. If that is what you meant, yes.

 

[Fernando Calvario]

Oh yes that's what I meant. I tend to forget the proper symbol

 

[haruspex]

Oh yes that's what I meant. I tend to forget the proper symbol

OK, so what do you get for the answer now?

 

[Fernando Calvario]

3.4496 m/s

 

[haruspex]

3.4496 m/s

The question asks for an angular speed, not a linear speed.

 

[Fernando Calvario]

*rad/s (some Freudian slip there hehe if the units is the only thing wrong)

 

[haruspex]

*rad/s (some Freudian slip there hehe if the units is the only thing wrong)

No, it's less than that.
What are you plugging into your equation in post #7 for α and θ_f?

 

[Fernando Calvario]

α = 5.00 * 10^-2 rad/s^2
θƒ = 162m
Oooh, I get it now I got the wrong θi. It should be 0

 

[haruspex]

θƒ = 162m

θƒ should be an angle, in radians. Remember - no linear quantities.

 

[Fernando Calvario]

θƒ = 23.672 rads (??)

 

[haruspex]

θƒ = 23.672 rads (??)

Looks about right. So what do you get now?

 

[Fernando Calvario]

≈1.5386 rad/s

 

[haruspex]

≈1.5386 rad/s

Much better. No need for so many digits, though.

 

[Fernando Calvario]

Much better. No need for so many digits, though.

I'll take note. Thank you