Conservation of angular momentum under central forces

[Saptarshi Sarkar]

Homework Statement

If a particle moves outward in a plane along a curved trajectory described by $r=a\theta$, where $\theta=\omega t$, where $a$ and $\omega$ are constants, then its

a) kinetic energy is conserved
b) angular momentum is conserved
c) total momentum is conserved
d) radial momentum is conserved

Relevant Equations

$p=m\dot r$
$L=mr^2\dot \theta$

I know that the force must be a central force and that under central forces, angular momentum is conserved. But I am unable to mathematically show if the angular and linear momentum are constants.

Radial Momentum
$p=m\dot r = ma\dot \theta=ma\omega$

Angular Momentum
$L=mr^2\dot\theta = ma^2\omega^3 t^2$

I am not sure if I am supposed to use the chain rule here, but I am getting a conserved linear momentum but a time-dependent angular momentum.

 

[PeroK]

$p=m\dot r = ma\dot \theta=ma\omega$

Angular Momentum
$L=mr^2\dot\theta = ma^2\omega^3 t^2$

I am not sure if I am supposed to use the chain rule here, but I am getting a conserved linear momentum but a time-dependent angular momentum.

Isn't that the answer for $L$?

For the other part, are you looking for radial momentum or total linear momentum?

 

[Saptarshi Sarkar]

Isn't that the answer for $L$?

For the other part, are you looking for radial momentum or total linear momentum?

I was thinking that the angular momentum must be independent of time as it looks like a trajectory under a central force. Is it not true?

For the other part, I did a mistake. Thanks for pointing it out. I need to consider the tangential velocity as well.

Total linear momentum = $p_{\text{tot}}=m(\dot r+r\dot\theta )=m(a\omega + a\omega^2 t)$

So, d should be the answer?

 

[PeroK]

I was thinking that the angular momentum must be independent of time as it looks like a trajectory under a central force. Is it not true?

For the other part, I did a mistake. Thanks for pointing it out. I need to consider the tangential velocity as well.

Total linear momentum = $p_{\text{tot}}=m(\dot r+r\dot\theta )=m(a\omega + a\omega^2 t)$

So, d should be the answer?

Yes. It's not a central force. You can calculate the acceleration if you want to. It won't be in the radial direction.

 

[vela]

Total linear momentum = $p_{\text{tot}}=m(\dot r+r\dot\theta )=m(a\omega + a\omega^2 t)$

The two components are perpendicular to each other. You can't just add them together as scalars.