Angular Velocity: Angular Kinematics Homework

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A cyclist traveling at 30 km/hr has a bike wheel diameter of 0.70 meters, leading to a need for calculating the angular velocity in rad/sec. The correct tangential speed is 8.33 m/s, not the initially considered 0.83 m/s. The relationship between linear speed (v) and angular speed (ω) is given by the equation v = ωr, where r is the radius of the wheel. By substituting the known values into this equation, the angular velocity can be determined. Understanding this relationship is crucial for solving similar angular kinematics problems.
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Homework Statement



A cyclist is traveling at a speed of 30km/hr. If the diameter of the bike wheel is 0.70 meter, what is the angular velocity of the wheels in rad/sec?


Homework Equations





The Attempt at a Solution



Hi, I have racked my brain trying ot work this out, I know the answer is 23.7 rad/sec
(but I don't know hoe to get there ;-(!
All I could come up with is 0.83m/s=speed and r=0.07m
 
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bionut said:

Homework Statement



A cyclist is traveling at a speed of 30km/hr. If the diameter of the bike wheel is 0.70 meter, what is the angular velocity of the wheels in rad/sec?


Homework Equations





The Attempt at a Solution



Hi, I have racked my brain trying ot work this out, I know the answer is 23.7 rad/sec
(but I don't know hoe to get there ;-(!
All I could come up with is 0.83m/s=speed and r=0.07m
Welcome to Physics Forums.

The speed of the bicycle is the tangential speed of the wheels (assuming no slipping). So, the question now becomes, how is the tangential speed related to the angular speed of a disc?
 
okay, not sure if I am thinking is on the right track, but would the angular speed be in reference to revolutions/time period? Where 1 revolution - 2(pi)r?
 
bionut said:
okay, not sure if I am thinking is on the right track, but would the angular speed be in reference to revolutions/time period? Where 1 revolution - 2(pi)r?
Almost, the definition of angular velocity (or speed) should be in your notes or textbook. Perhaps you could look it up.
 
okay, all it says is angular speed = angular distance / change in time

If the total distance in 30 km = 3000m

r= 0.7m

I really just don't get it sorry... I am sure its very basic and I am missing something here... ill kick myself when I find out that's for sure !
 
Okay, thankyou for your help... but I am back to sqaure 1 lol... gawd !

okay if v=wr and s=2pi and the displacmeent is 2pi (6.8 rad) X 0.70 = 4.40, and v=30km/hr , is the tangential velcoity 0.83m/s
 
bionut said:
Okay, thankyou for your help... but I am back to sqaure 1 lol... gawd !

okay if v=wr and s=2pi and the displacmeent is 2pi (6.8 rad) X 0.70 = 4.40, and v=30km/hr , is the tangential velcoity 0.83m/s
Not quite. 30 km/hr is 8.33... m/s. So, the equation you need is

v = \omega r

Now, you have v (8.33 m/s) and you can work out the radius, r. And you need to work out \omega. Do you think you can do that?
 
... oh dear... it is 8.3m/s ... no wonder I have spent 1/2 the day try to work it out ... lol... told you I would kickmyself... instead I think i need paperbag lol...
 
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