The wheels of a skateboard roll without slipping as it accelerates at 0.40 m/s2 down an 85-m-long hill. Part A If the skateboarder travels a

Question

The wheels of a skateboard roll without slipping as it accelerates at 0.40 m/s2 down an 85-m-long hill. Part A If the skateboarder travels at 1.8 m/s at the top of the hill, what is the average angular speed of the 2.6-cm-radius wheels during the entire trip down the hill

in progress 0
Gerda 3 years 2021-08-26T13:29:05+00:00 1 Answers 14 views 0

Answers ( )

  1. Adela
    0
    2021-08-26T13:30:44+00:00
    Reply

    Answer:

    average angular speed = 196.92 rad/s

    Explanation:

    Given data

    accelerates down = 0.40 m/s²

    down hill travel = 85-m

    travels top of hill = 1.8 m/s

    radius wheels =  2.6 cm

    Solution

    We will apply here equation of motion that is express as

    v^{2}-u^{2} = 2as    ……………1

    here v is final velocity  and u is final velocity  and s is dispalement

    so put here value we get first final velocity

    v^{2} = 1.8^{2}  + 2 × 0.40 × 85

    solve it we get

    v^{2}  = 71.24


    v = 8.44 m/s

    and

    initial angular speed is express as

    initial angular speed ω = \frac{u}{r} …………2

    put here value

    initial angular speed ω = \frac{1.8}{2.6 \times 10^{-2}}

    initial angular speed ω = 69.23 rad/s  

    and

    final angular speed ω = \frac{v}{r}    …………..3

    put here value

    final angular speed ω =  \frac{8.44}{2.6 \times 10^{-2}}  

    final angular speed ω = 324.61 rad/s  

    so now we get  average of angular speed that is

    average angular speed = ( 69.23 + 324.61  ) ÷ 2

    average angular speed = 196.92 rad/s

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )