Riding a bicycle around a tree w/ constant acceleration

AI Thread Summary
The discussion revolves around calculating the final angular speed of a bicycle rider traveling around a baobab tree with a circumference of 43.0 m, starting from rest and experiencing a constant angular acceleration of 5.00 × 10–2 rad/s² over a distance of 162 m. The participants emphasize the importance of using angular quantities rather than linear ones, leading to the correct application of the equation ωf² = ωi² + 2α(θf - θi). After clarifying the proper symbols and units, the final angular speed is determined to be approximately 1.5386 rad/s. The conversation highlights the parallels between linear and angular motion, underscoring the need for accurate conversions between the two. Overall, the thread serves as a learning experience for applying angular motion equations correctly.
Fernando Calvario
Messages
13
Reaction score
1

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
 
Physics news on Phys.org
Fernando Calvario said:
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?
 
haruspex said:
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
 
Fernando Calvario said:
vf ^2 = vi^2 + 2aΔx
Right. Try converting that to angular entities.
 
So I did this: ωf = √2(5*10^-2 rad/s^2)(162m-43m). I got 3.446 m/s
 
Fernando Calvario said:
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.
 
haruspex said:
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)
 
Fernando Calvario said:
ωf2 = ωi2 + 2a (θf - θi)
What is a there? Angular acceleration is usually written α. If that is what you meant, yes.
 
Oh yes that's what I meant. I tend to forget the proper symbol
 
  • #10
Fernando Calvario said:
Oh yes that's what I meant. I tend to forget the proper symbol
OK, so what do you get for the answer now?
 
  • #11
3.4496 m/s
 
  • #12
Fernando Calvario said:
3.4496 m/s
The question asks for an angular speed, not a linear speed.
 
  • #13
*rad/s (some Freudian slip there hehe if the units is the only thing wrong)
 
  • #14
Fernando Calvario said:
*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 θf?
 
  • #15
α = 5.00 * 10^-2 rad/s^2
θƒ = 162m
Oooh, I get it now I got the wrong θi. It should be 0
 
  • #16
Fernando Calvario said:
θƒ = 162m
θƒ should be an angle, in radians. Remember - no linear quantities.
 
  • #17
θƒ = 23.672 rads (??)
 
  • #18
Fernando Calvario said:
θƒ = 23.672 rads (??)
Looks about right. So what do you get now?
 
  • #19
≈1.5386 rad/s
 
  • #20
Fernando Calvario said:
≈1.5386 rad/s
Much better. No need for so many digits, though.
 
  • #21
haruspex said:
Much better. No need for so many digits, though.
I'll take note. Thank you
 
Back
Top