Can L be conserved if its magnitude is conserved?

[subzero0137]

Problem: Consider a system for which Newton's second law is $$ \frac {d \vec v}{dt} = - [ \frac {h(r)h'(r)}{r} + \frac {k}{r^3} ] \vec r- \frac {h'(r)}{r} \vec L $$ where k is a constant, h(r) is some function of r, h'(r) is its derivative and L = r x v is the angular momentum. Show that $$ \frac {d \vec L}{dt} = - \frac {h'(r)}{r} \vec r × \vec L$$ and use this equation to prove that L is not generally conserved, but its magnitude L is conserved.

Attempt: I've done the first part of the question, but I don't know how I should go about showing that L is not conserved but its magnitude is conserved. Any hints would be appreciated.

 

[Dick]

Problem: Consider a system for which Newton's second law is $$ \frac {d \vec v}{dt} = - [ \frac {h(r)h'(r)}{r} + \frac {k}{r^3} ] \vec r- \frac {h'(r)}{r} \vec L $$ where k is a constant, h(r) is some function of r, h'(r) is its derivative and L = r x v is the angular momentum. Show that $$ \frac {d \vec L}{dt} = - \frac {h'(r)}{r} \vec r × \vec L$$ and use this equation to prove that L is not generally conserved, but its magnitude L is conserved.

Attempt: I've done the first part of the question, but I don't know how I should go about showing that L is not conserved but its magnitude is conserved. Any hints would be appreciated.

The magnitude squared is given by the dot product of L with itself. Can you show the time derivative of that is 0?

 

[subzero0137]

So |L|^2 = (r×v)⋅(r×v) = (r⋅r)(v⋅v) - (v⋅r)(v⋅r) = |r|^2 |v|^2 right? But how would I show the time derivative of this to be 0? $\frac {dL}{dt} = r \frac {dv}{dt} + v \frac {dr}{dt}$, but how do I make this equal 0?

 

[Dick]

So |L|^2 = (r×v)⋅(r×v) = (r⋅r)(v⋅v) - (v⋅r)(v⋅r) = |r|^2 |v|^2 right? But how would I show the time derivative of this to be 0? $\frac {dL}{dt} = r \frac {dv}{dt} + v \frac {dr}{dt}$, but how do I make this equal 0?

You want to show the time derivative of $L \cdot L$ is zero. Use the product rule and your given expression for dL/dt. Can you tell me why dL/dt must be perpendicular to L?