A 9.29 kg particle with velocity is at x = 7.86 m, y = 9.84 m. It is pulled by a 9.78 N force in the negative x direction. About the origin,

Question

A 9.29 kg particle with velocity is at x = 7.86 m, y = 9.84 m. It is pulled by a 9.78 N force in the negative x direction. About the origin, what are (a) the particle’s angular momentum, (b) the torque acting on the particle, and (c) the rate at which the angular momentum is changing?

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RI SƠ 5 years 2021-08-14T10:06:14+00:00 1 Answers 15 views 0

Answers ( )

  1. thnahdat
    0
    2021-08-14T10:07:51+00:00
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    Explanation:

    The question didn’t state the velocity in terms of i and j but let take it as v= (6.0m/s)i-(7.0m/s)j

    m=9.29kg

    x=7.86m

    y=9.84m

    F=9.78N

    a. To find particle’s angular momentum:

       l=m(r*v)    where r= x+y  r=(7.86i+9.84j)

       l=9.29[(7.86i+9.84j)(6i-7j)]\\l=9.29*[-55.02ij+59.04]\\l=1059.6k kgm^{2} /s

    b.To find torque:

       r=r*F\\r=(7.86i+9.84j)m*(-9.78)N\\r=9.29kg*[-96.24ij]Nm\\r=(+96.24k)Nm

    c.

    The Newton’s 2nd law of motion said that the rate change of momentum is equal to the net torque,

    so Δl/Δt = (+96.24k)Nm

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