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Rolling Motion

  1. Jun 8, 2009 #1
    1. The problem statement, all variables and given/known data
    Assume that there is no slipping of the surfaces in contact during the rolling motion.

    A motorcycle accelerates uniformly from rest and reaches a linear speed of 22.0 m/s in a time of 9s. The radius of each tire is 0.280 meters. What is the magnitude of the angular acceleration of each tire?

    2. Relevant equations

    really not sure what im doing.

    3. The attempt at a solution
    In the book it says that when there is no slipping, that tangential velocity = linear velocity. And since [tex]\alpha[/tex] = [tex]\Delta\omega[/tex]/[tex]\Deltat[/tex] , took 22 and divided it by 9 seconds..
  2. jcsd
  3. Jun 8, 2009 #2
    and then I multiplied it by the radius .280

    because tang. acceleration = r * angular acceleration
  4. Jun 8, 2009 #3


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    Yes. Do you understand WHY this is the case, though? It's saying that if your tires have perfect grip, then for every centimetre that the tires advance, the vehicle advances by the same distance. Can you see that if you didn't have perfect grip/traction, then the tires would slide relative to the road, and so the tangential velocity of the tires would NOT necessarily have to be equal to the linear velocity of the vehicle? For example, imagine you did a "burn" in which you hit the brakes and the accelerator at the same time. Your tires would be spinning in place quite fast (with some non-zero velocity), whereas your vehicle would be going nowhere (zero velocity). This is becase your tires would be slipping (there would be relative motion between the tires and the road...the two surfaces would be sliding past each other as you "burned" rubber). I hope this explains to you why friction is necessary to prevent slipping and enable a vehicle to move forward when its tires spin. It is the reaction force of friction from the road on the tires that propels the vehicle forward.

    Umm yeah. The problem with that is that omega is not equal to 22 m/s, so plugging in 22 m/s for omega makes no sense. What you have been given is the linear velocity, v, of the bike (or equivalently, the rotational velocity of of a point on the edge of a tire. What is the relationship between that velocity and the ANGULAR velocity of the tire (omega)?

    Hint: As a more direct approach to the problem, what is the relationship between the linear acceleration of the bike (which you can calculate using the data you have been given), and the angular acceleration of its tires?
    Last edited: Jun 8, 2009
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