Understanding Engineering Mathematics

is a vector tangential to the curve, pointing in the direction of motion of the particle and

with magnitude equal to the magnitude of the velocity of the particle. That is,

dr

dt

is the

vectorvelocityof the particle:

v=

dr

dt

Similarlya=

d^2 r

dt^2

is theaccelerationof the particle.

Problem 11.13

Find the velocity, acceleration and kinetic energy of the particle massm

whose position vector at timetis:

r.t/=2cos!tiY2sin!tj

Verify that for such a particle v·r=0.

The velocity is

v=

dr

dt

=− 2 ωsinωti+ 2 ωcosωtj

From this we find

v·r= 4 ω(−sinωti+cosωtj)·(cosωti+sinωtj)

= 4 ω(−sinωtcosωt+cosωtsinωt)= 0

The acceleration is

a=

d^2 r

dt^2

=

dv

dt

=− 2 ω^2 cosωti− 2 ω^2 sinωtj

=−ω^2 r

The kinetic energy is

1

2

mv^2 =

1

2

m((− 2 ωsinωt)^2 +( 2 ωcosωt)^2 )

=

1

2

m· 4 ω^2

= 2 mω^2

Exercises on 11.13

1. Show that ifeis a vector of constant magnitude, then

e·

de

dt

= 0