What is the Angular Deceleration of a Bicycle Wheel?

Click For Summary
SUMMARY

The discussion focuses on calculating the angular deceleration of a bicycle wheel based on observed drop heights from the wheel during repairs. The initial and subsequent velocities of the drops were calculated as 2.6674 m/s and 2.4887 m/s, respectively, corresponding to heights of 36.3 cm and 31.6 cm. The time for one complete cycle was determined to be 0.9422 seconds for the first drop and 1.0099 seconds for the second. The participant seeks guidance on applying kinematic formulas to derive angular deceleration, emphasizing the need to convert tangential quantities into angular quantities.

PREREQUISITES
  • Understanding of kinematic equations for linear motion
  • Knowledge of angular motion concepts, including angular velocity and angular acceleration
  • Familiarity with unit conversions between linear and angular measurements
  • Basic grasp of gravitational effects on projectile motion
NEXT STEPS
  • Learn how to apply the kinematic equation for angular acceleration: α = (ω_f - ω_i) / t
  • Study the relationship between tangential and angular quantities, specifically how to convert linear speed to angular speed using the formula: ω = v / r
  • Explore the use of the maximum height formula in the context of angular motion to derive initial velocities
  • Investigate the implications of angular deceleration on the motion of rotating bodies
USEFUL FOR

Students studying physics, particularly those focusing on rotational dynamics and kinematics, as well as educators seeking to clarify concepts related to angular motion and deceleration.

cdubsean
Messages
11
Reaction score
0

Homework Statement



A bicycle is turned upside down while its
owner repairs a flat tire. A friend spins the
other wheel of radius 0.4 m and observes that
drops of water fly off tangentially. She mea-
sures the height reached by drops moving ver-
tically. A drop that breaks loose from the tire
on one turn rises 36.3 cm above the tangent
point. A drop that breaks loose on the next
turn rises 31.6 cm above the tangent point
(the angular speed of the wheel is decreas-
ing).

Find the angular deceleration of the wheel.
The acceleration of gravity is 9.8 m/s2 . As-
sume the angular deceleration is constant.
Answer in units of rad/s2.

Homework Equations



Max height = (Vi^2)/2g
2pi(r)
Angular Velocity


The Attempt at a Solution



Given what I know I converted units
36.3 cm to .363m
31.6 cm to .316m

Then used max height formula to determine the Vi

Vi1 = 2.6674 m/s
Vi2 = 2.4887 m/s

and with this I can find the time buy using the circumference and the velocities to determine the times.

T1= .9422s for one complete cycle
T2= 1.0099s for one complete cycle

And with this I need to use angular acceleration formula to get this, but here is where I am lost and fell like I am going in wrong direction. Can I get some guidance into what I should do next. I know that I might have to use this to find tangent line which between these two to find acceleration...in this cause deceleration. But I am lost.
 
Physics news on Phys.org
cdubsean said:
and with this I can find the time buy using the circumference and the velocities to determine the times.

T1= .9422s for one complete cycle
T2= 1.0099s for one complete cycle
Instead of this, which assumes constant speed over the course of each cycle, trying using a different kinematic formula that gives you the acceleration directly.

How do you convert tangential quantities (distance, speed, acceleration) to angular quantites?
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
12K
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 121 ·
5
Replies
121
Views
12K
  • · Replies 2 ·
Replies
2
Views
5K