Totally confused acceleration and velocity of a particle in vectors(cross product)

[CmbkG]

Totally confused! acceleration and velocity of a particle in vectors(cross product)

Homework Statement

The acceleration of a particle is given by a=vx(cxr) where r is the position, v is the velocity and c is a constant.

Show that the following are constants
(a)|v|
(b)c.(rxv)
(c)c.v-1/2|cxr|^2

The Attempt at a Solution

a=c(v.r)-r(v.c)

d/dta=d/dt[c(v.r)-r(v.c)]

 

[Dick]

Something is constant if its time derivative is zero. |v| is constant if v.v is constant. Take d/dt(v.v). Remember d/dt(v)=a.

 

[CmbkG]

Thanks for the help.

Ive got part (a) and it equals zero. Its just I am not too sure on what way to even start part (b) or (c). Should i use the cartesian components or how do i find out what r and c actually equal?

 

[weejee]

(b) d(c.(rxv))/dt = c.(vxv) + c.(rxa) = c.(rxa)
Try to show that this equals zero.

(c) d(c.v)/dt = c.a
d(1/2|cxr|^2)/dt = (cxr).(cxv)
Try to show that these two equals each other.

 

[Dick]

How did you get a)? A very similar technique will get you b) and c). Start by taking d/dt of the expressions. For c) you might find the identity (axb).c=a.(bxc) useful.

 

[CmbkG]

Thanks for both your help there, appreciate it alot.

Just wondering though weejee, how did you get d(c.(rxv))/dt to be equal to c.(vxv)+c.(rxa)? where did the vxv and the rxa come from?

 

[weejee]

Your welcome.

The differentiation can act on either 'r' or 'v'. If it acts on 'r' it gives c.(vxv) and if it acts on 'v' it leads to c.(rxa).

 

[CmbkG]

ahh, rite, i see what you mean now. Thanks alot.

 

[Dick]

Your welcome.

The differentiation can act on either 'r' or 'v'. If it acts on 'r' it gives c.(vxv) and if it acts on 'v' it leads to c.(rxa).

It's called the 'product rule'.