A 5.00-kg particle starts from the origin at time zero. Its velocityasa functionoftimeisgivenbyv = (6t 2i+ 2t j) where v is in meters per second and t is in seconds. (a) Find its position vector r as a function of time (b) its acceleration as a function of time,(c) the net force exerted on the particle as a function of time, (d) the net torque about the origin exerted on the particle as a function of time,(e)the angular momentum of the particle as a function of time,(f)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.Hint: (a) v = dr/dt and dr = vdt, r =òvdt (integrate v and get r),(b) a=dv/dt, © F=ma,(d) t = r x F (cross product) (e) L= r x mv(cross product) (f) KE=(1/2)mv2(Square v vector to get v2,i2 =j2 =1),(g) Power = first derivative of KE

Solution given thatm=5 kgu=6t2i +2tjms

Answered: A 5.00-kg particle starts from the…

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter11: Angular Momentum

Section: Chapter Questions

Problem 89AP: A solid cylinder of mass 2.0 kg and radius 20 cm is rotating counterclockwise around a vertical axis...

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A 5.00-kg particle starts from the origin at time zero. Its velocityasa functionoftimeisgivenbyv = (6t 2i+ 2t j) where v is in meters per second and t is in seconds. (a) Find its position vector r as a function of time (b) its acceleration as a function of time,(c) the net force exerted on the particle as a function of time, (d) the net torque about the origin exerted on the particle as a function of time,(e)the angular momentum of the particle as a function of time,(f)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.

Hint: (a) v = dr/dt and dr = vdt, r =òvdt (integrate v and get r),(b) a=dv/dt, © F=ma,(d) t = r x F (cross product) (e) L= r x mv(cross product) (f) KE=(1/2)mv2(Square v vector to get v2,i2 =j2 =1),(g) Power = first derivative of KE

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