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A ball is spun with angular acceleration a = 6t2 – 2t where t is in second and a is in rads-1. At t = 0, the ball has angular velocity of 10 rads-1 and angular position of 4 rad. The most appropriate expression for the angular position of the ball is:

Option 1 -

3 2 t 4 − t 2 + 1 0 t

Option 2 -

t 4 2 − t 3 3 + 1 0 t + 4

Option 3 -

2 t 4 3 − t 3 6 + 1 0 t + 1 2

Option 4 -

2 t 4 − t 3 2 + 5 t + 4

0 2 Views | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    2 months ago
    Correct Option - 2


    Detailed Solution:

    a = 6t2 – 2t,        of t 20, w = 10 rad/sec

    , q = 4 rad

    α=dωdt

    ⇒∫10ωdω=∫t−0tαdt

    ⇒ω−10=∫0t (6t2−2t)dt

    ω=dθdt

    dθ=ω dt

    ∫4θdθ=∫0t (2t3−t2+10)dt

    = t42−t33+10t+4

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A
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