Answer to Question #154769 in Classical Mechanics for Cove

Question #154769
  1. Show that the rate of change of the total angular momentum of a system of particles is equal to the resultant torque exerted by all external forces which act on the system. 
1
Expert's answer
2021-01-11T11:41:22-0500

Let the particle of mass m, is moving with velocity v\overrightarrow{v} then linear momentum of the particle will be p\overrightarrow{p} .and particle is at a distance r\overrightarrow{r} from the origin.

Let the angular momentum of the particle is L.\overrightarrow{L}.



So, angular momentum of the particle will be L=r×p\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}

L=rpsinθn^\Rightarrow \overrightarrow{L}=|\overrightarrow{r}||\overrightarrow{p}|\sin\theta \hat{n}

where θ\theta is the angle between r\overrightarrow{r} and p\overrightarrow{p} and n^\hat{n} is the unit vector perpendicular to r\overrightarrow{r} and p\overrightarrow{p}

Now, taking the time derivative of L=r×p\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}

dLdt=drdt×p+r×dpdt\Rightarrow \frac{d\overrightarrow{L}}{dt}=\frac{d \overrightarrow{r}}{dt}\times \overrightarrow{p}+\overrightarrow{r}\times \frac{{d\overrightarrow{p}}}{dt}

from newton's second law of motion,ΣF=dpdt\Sigma \overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}

Hence, dLdt=r×F\frac{d\overrightarrow{L}}{dt}=\overrightarrow{r}\times \overrightarrow{F}

But we know that τ=r×F\overrightarrow{\tau}= \overrightarrow{r}\times \overrightarrow{F}

Hence, τ=dLdt\overrightarrow{\tau}=\frac{d\overrightarrow{L}}{dt}


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