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?

Question

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Respuesta :

fahadisahadam fahadisahadam · Answer 1 · 2020-03-26T04:21:26+01:00

Explanation:

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

[tex]m=9.29kg[/tex]

[tex]x=7.86m[/tex]

[tex]y=9.84m[/tex]

[tex]F=9.78N[/tex]

a. To find particle's angular momentum:

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

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

b.To find torque:

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

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