Anonymous
Anonymous asked in Science & MathematicsPhysics · 3 months ago

Mountain Bike?

Starting from rest at t=0sec, a tire on a mountain bike has a constant angular acceleration. When t= 4 sec, the angular velocity of the wheel is 7 rad/sec. The angular acceleration continues until t= 19 sec, after which the angular velocity remains constant. The tire has a radius of 22cm.

1)What is the angular acceleration at t= 19 sec?

α =

 rad/sec2

2)What is angular velocity at t= 19 sec?

ω =

 rad/sec

3)What is the total angular displacement of the tire from t=0sec to t=19 sec?

θ =

 rad

4)What would be the total displacement of a small stone stuck in the tread of the tire as it goes around from t=0sec to t=19 sec?

Δs =

 m

5)What would be the linear displacement of the tire's axle as it moves forward from t=0sec to t=19 sec?

Δx =

 m

2 Answers

Relevance
  • NCS
    Lv 7
    3 months ago
    Favourite answer

    1) α = Δω / Δt = 7 rad/s / 4s = 1.75 rad/s²

    2) α = 0

    for t ≥ 19 s

    the way I read it. It's possible they want 1.75 rad/s².

    3) Θ = ½αt² = ½ * 1.75rad/s² * (19s)² = 316 rads

    4) s = rΘ = 0.22m * 316rads = 69.5 m

    5) d = s / 2π = 11.1 m

    Hope this helps!

  • 3 months ago

    NCS has given you good answers but I disagree on some interpretation

    1.  Undefined. It is 7/4 BEFORE 19 and 0 AFTER 19 .  It is undefined AT 19.

    2.  w = at = 7/4 *19 = 33.25 rad/s

    3.  theta = average w * t = 1/2 * 33.25 * 19 = 316 radians

    5.  I will do this first.  The distance that the circumference has moved is angle * radius.  = 1/2 * 33.25*19*0.22 ~= 69.5 m.  Unless the tyre slips on the ground the axle moves forward by the same distance that the circumference lays along the roadway.

    4.  This is hard.  At this instant where is the stone relative to the axle? and where was it at the beginning.  It was 0.22 m below the axle at the beginning.  It has done 50 complete turns PLUS 1.75 radians.

    The stone is now cos(1.75)*0.22 = - 0.04m below the axle and sin(1.75)*0.22 = 0.22 m behind the axle.  Giving a displacement of 69.5-0.22 forward and 0.22 + 0.04 above its original position.

    This will NOT be the answer the computer is expecting but it IS the answer to the question as asked.

    They asked for DISPLACEMENT which is a vector of where the stone is now - where it was at the beginning.  This part of the question is not well worded.

Still have questions? Get answers by asking now.