Calculating Velocity of a Wheel at the Top Using Acceleration and Time

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A cyclist accelerates from rest at 1.10 m/s² for 3.10 seconds, resulting in a linear velocity of 3.41 m/s. The point on the rim of the wheel at the top moves at twice this speed, reaching 6.82 m/s, while the point in contact with the ground remains stationary. The diameter of the wheel is not necessary for this calculation, as the focus is on linear rather than rotational speed. The discussion emphasizes understanding the relationship between linear and rotational motion. Further assistance is offered for additional problems.
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1. Homework Statement

A cyclist accelerates from rest at a rate of 1.10 m/s2. How fast will a point on the rim of the tire (diameter = 79.7 cm) at the top be moving after 3.10 s?


2. The attempt at a solution:

i just tried this one...

a_avereage= Δw/Δt so a_avereage= ( w_2 - w_1 ) / Δt

if we set for w_2 we get w_2 = W_1 + (a_avereage)(Δt)
with w_1=0 the response for w_2 is 1.358 rad/s but that was wrong

can anyone help me how to ge the velocity of the wheel at the top t 3.10 s?
 
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The first thing you need to figure out is how fast the bike is moving after 3.10 s. What's that? That's the same speed that the hub of the wheel is moving at. Now think about it. A point on the rim of the wheel that is in contact with the road is not moving relative to the road (if the bike isn't skidding). Right? The means the rotational speed of a point on the wheel is the same as the linear speed of the bike but in the opposite direction so they can cancel. Try drawing a picture. What happens at the top of the wheel? That's the easy way to think about it. You don't need the diameter of the wheel.
 
i don't undertand whau told me...

i need to know how to calculate that velocity?
 
Naldo6 said:
i don't undertand whau told me...

i need to know how to calculate that velocity?

Yes, you do, whether you do it the way I suggested or not. The bike accelerates at 1.10m/s^2 for 3.10s. What's its velocity?
 
ok, but how i calculate that velocity?... i tried in the above way but that was wrong.
 
It's a linear acceleration problem so far. Forget the rotational formulas.
 
ok so v= v_i + at
where v_i = 0
so v= (1.10m/s^2)(3.10s)=3.41 m/s right?...
 
That's a good start. Yes. So what's the rotational speed of points on the rim of the wheel? How fast is the rim of the tire spinning? In m/s, not radian/s.
 
is at the top 2 times the velocity and at the bottom zero?
 
  • #10
Sure it is. The bottom of the wheel is at rest, so the top of the wheel is moving at 2*3.41m/s forward. You've got the picture, right? See? No diameter needed.
 
  • #11
ty a lot...

can u help m ewith other problems i have doubt?...

i will post in new topics...
 
  • #12
I'll try. You're welcome.
 
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