Turning wheel, simply angular momentum

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Homework Help Overview

The problem involves a bicycle with a specified tire diameter coasting at a given speed, focusing on the angular speed of the tire and the speed of a dot painted on the tire at different heights above the road. The subject area includes concepts of angular momentum and relative motion.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to calculate the angular speed of the tire and the speed of the dot at different heights, questioning why the speed is not simply the same as the bicycle's speed for all cases.
  • Some participants question the relationship between the dot's speed and the bicycle's speed, particularly at different heights, and explore the concept of relative motion.
  • Others suggest considering the vector sum of the center of mass speed and the dot's speed relative to the center of mass.

Discussion Status

Participants are actively discussing the relationship between the speeds involved, with some providing clarifications on the vector nature of the problem. There is an ongoing exploration of how to correctly calculate the speed of the dot at different heights, and some guidance has been offered regarding the vector addition of velocities.

Contextual Notes

There is a focus on understanding the relative speeds and the implications of the dot's position on the tire, with participants acknowledging the complexity of the motion as the wheel rolls. The discussion reflects an attempt to reconcile different interpretations of the problem setup.

Ground State
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Homework Statement



A bicycle with 0.8 m diameter tires is coasting on a level road at 5.6 m/s. A small dot has been painted on the rear tire. What is:
a) the angular speed of the tire
b) the speed of the dot when it is 0.8 m above the road in m/s?
c) the speed of the dot when it is 0.4 m above the road in m/s?


The Attempt at a Solution



a)
ω=v/r
ω=5.6/0.4
ω=14

for (b) and (c), can someone help me understand why the answer is not 5.6 m/s for both of them?
 
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Ground State said:
for (b) and (c), can someone help me understand why the answer is not 5.6 m/s for both of them?
In this case, they want the speed of the dot relative to the road (relatvie to any point on the road).
 
rcgldr said:
In this case, they want the speed of the dot relative to the road (relatvie to any point on the road).

But when it is 0.8 m above the road, and hence, the full diameter of the wheel, is the speed not tangentially in the same direction as the road, and therefore 5.6 m/s? (It isn't, I know this answer is wrong but am not sure why). Is it because it will have the cumulative linear speed of the moving car and angular speed of the turning wheel? Does that mean that the answer for (c) will simply be answer from (a)?
 
See picture. The centre of the tyre moves with 5.6 m/s to the right. The speed of the perimeter of the tyre is also 5.6 m/s, but the velocity of the dot with respect to the CM changes direction as the wheel rolls.

The velocity of the dot with respect to the road is the vector sum of the velocity of CM and the velocity of the dot with respect to the CM . And the speed of the dot is the magnitude of velocity.

ehild
 

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ehild said:
The velocity of the dot with respect to the road is the vector sum of the velocity of CM and the velocity of the dot with respect to the CM . And the speed of the dot is the magnitude of velocity.

ehild

So am I understanding your explanation correctly by submitting the following?

(b)
velocity with respect to the road = velocity of CM + velocity of the dot with respect to the CM
velocity with respect to the road = 5.6 m/s + 5.6 m/s
velocity with respect to the road = 11.2 m/s

(c) (arranging vectors tip to tail and taking resultant)
= (5.6 m/s)^2 + (5.6 m/s)^2
= 7.92 m/s
 
It is correct now. And what is the speed of the dot when it is at the bottom of the wheel? ehild
 

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