A 5.00-kg particle starts from the origin at time zero. Its velocity as a functi

Question

A 5.00-kg particle starts from the origin at time zero. Its velocity as a function of time is given by v->= 6t^2 i^ + 2t j^ where v-> is in meters per second and t is in seconds. (a) Find its position as a function of time. (b) Describe its motion qualitatively. Find (c) its acceleration as a function of time, (d) the net force exerted on the particle as a function of time, (e) the net torque about the origin exerted on the particle as a function of time, (f) the angular momentum of the particle as a function of time, (g) the kinetic energy of the particle as a function of time, and (h) the power injected into the system of the particle as a function of time.

Explanation / Answer

v = 6t^2 i + 2t j

a) position = integral ( v dt)

= integral ( 6t^2 i + 2t j ) dt

= 2t^3 i + t^2 j

c) a = dv/dt = 12 t i + 2 j

d) net force = ma = 5(12t i + 2 j ) = 60t i +10 j

e) net torque about origin = F*r = 60t i +10 j * (sqrt(60^2 + 10^2)) = 3649t i + 608.2 j

f) angular momentum = m(r X v) = 5 [ i j k ]

[ 2t^3 t^2 0]

[ 6t^2 2t 0]

= 5 [ i(0 -0 ) - j(0 - 0) +k(4t^4 - 6t^4)]

= -10t^4 k

g) KE = 0.5mv^2 = 0.5*5*(6t^2i + 2t j )^2

= 2.5*(36t^4i + 4t^2j)

= 90t^4 i + 10t^2 j

h) P = F*v = 60t i + 10 j . (6t^2i + 2t j )

= 360t^3 i + 20t j

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