Angular Displacement: Solving for a Mountain Bike Wheel

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Homework Help Overview

The problem involves calculating the angular displacement of a mountain bike wheel under constant angular acceleration, transitioning to constant angular velocity. The original poster describes the scenario where the wheel starts from rest and accelerates until a specified time, after which it maintains a constant speed.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss dividing the problem into two parts: one with constant angular acceleration and the other with constant angular velocity. There are attempts to calculate angular acceleration and displacement using different methods, including the use of average velocity.

Discussion Status

Participants are actively engaging with the problem, offering corrections and alternative approaches. Some guidance has been provided regarding the calculation of angular acceleration and the need to determine the correct speed for the constant velocity portion. There is ongoing exploration of the calculations and assumptions involved.

Contextual Notes

There are indications of confusion regarding the timing of the angular velocity measurements and the application of formulas for angular displacement. Participants are also noting the constraints of the problem, including the limited number of attempts to submit answers.

SnowOwl18
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Alright, this problem is really starting to bug me.

----Starting from rest at t0 = 0s, a wheel on a mountain bike has a constant angular acceleration. When t = 5.15s, the angular velocity of the wheel is +7.66rad/s. The angular acceleration continues until t = 13.6s, after which time the angular velocity remains constant. What is the angular displacement of the wheel in the time interval from t = 0 to 28.0s?----

I divided it into two parts, since angular acceleration is constant from t=0s to t=13.6s, and isn't for the remaining portion. So for that first part, I calculated the angular acceleration by doing change in angular velocity or change in time...7.66rad/s / 13.6s = 0.562 rad/s^2. And then I solved for Theta = 0.5 A T^2 ...and I got 52.088 rad. And then for the second part I found the acceleration over the remaining 14.4s and got an acceleration (through the same process) of 0.532 rad/s^2...and then I solved for Theta of that portion and then added my two numbers to get 107.24 rad. But it's wrong. Any guidance would be much appreciated. Thanks! :)
 
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SnowOwl18 said:
I divided it into two parts, since angular acceleration is constant from t=0s to t=13.6s, and isn't for the remaining portion. So for that first part, I calculated the angular acceleration by doing change in angular velocity or change in time...7.66rad/s / 13.6s = 0.562 rad/s^2. And then I solved for Theta = 0.5 A T^2 ...and I got 52.088 rad.
You made a mistake in calculating the acceleration: 7.66 rad/s is the angular speed at t = 5.15 s, not t = 13.6 s. So the change in time is 5.15 s.

And then for the second part I found the acceleration over the remaining 14.4s and got an acceleration (through the same process) of 0.532 rad/s^2...and then I solved for Theta of that portion and then added my two numbers to get 107.24 rad.
For the second part, the acceleration is zero. They tell you that the speed is constant.
 
Last edited:
O ok...so for the first part (t=0s to t=5.15s) 7.66/10.3= 0.738 rad. I had thought the acceleration would be zero because of constant velocity (now if I only followed my intuition!)...so that would mean I just multiply the speed by the time to get distance. So if I multiply 7.66 by the remaining time, 22.85s, and get 175.769 rad...and then add the 0.738 rad i found before...would that be the correct way to find the answer?
 
First, let me apologize for telling you to use the average velocity to find the displacement. You can't use that method since you don't have the final speed at the end of the first part of the motion. D'oh! I must have been sleeping. (I removed that advice from the previous post.)
SnowOwl18 said:
O ok...so for the first part (t=0s to t=5.15s) 7.66/10.3= 0.738 rad. I had thought the acceleration would be zero because of constant velocity (now if I only followed my intuition!)...so that would mean I just multiply the speed by the time to get distance. So if I multiply 7.66 by the remaining time, 22.85s, and get 175.769 rad...and then add the 0.738 rad i found before...would that be the correct way to find the answer?
Here's what you do. As you stated earlier, treat the motion in two parts (1) accelerated (from t = 0 to 13.6) and (2) constant speed (from t = 13.6 to 28.0).

For part 1, find the acceleration. (See my previous post.) Then use that acceleration to do two things: find the displacement (using \theta = 1/2 \alpha t^2) over the entire interval, and find the find speed at the end of the interval (using \omega_f = \alpha t).

For part 2, you have the speed and the time so use \theta = \omega t.

Now just add the two displacements.
 
For part two I did 7.66 rad/s x 14.4s = 110.304 rad

Part one...a= deltav/deltat...7.66rad/s / 13.6s = 0.563 rad/s^2 ...
.5 x 0.563 x 13.6^2 = 52.07 rad

When added together, I get 162.4 rad total...but the program says I'm wrong...and unfortunately, I have only one try left. Sorry to keep bothering you, but do you see another mistake? Thanks so much for your help thus far...I don't know why I keep messing this up. :/
 
You are still making errors in both parts.
SnowOwl18 said:
For part two I did 7.66 rad/s x 14.4s = 110.304 rad
7.66 rad/s is only the speed at t=5.15s. It is not the speed you need for the "constant speed" part of the motion (part 2). You have to figure out that speed. See my last post for how to figure out that speed using the acceleration from part 1.
Part one...a= deltav/deltat...7.66rad/s / 13.6s = 0.563 rad/s^2 ...
.5 x 0.563 x 13.6^2 = 52.07 rad
See my first post. You are using the wrong time in calculating the acceleration: \alpha = \Delta \omega/\Delta t = (7.66)/(5.15) [{rad}/s^2].
 

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