Question

prove that the rate of change angular momentum of a system of particles about a reference point is equal to the net torqe acting on the system​

Answer 1

First let's derive expressions for torque τ and angular momentum L.

Consider a body rotating about an axis. When an external torque is applied on the body, its angular velocity changes, hence the body gets angular acceleration. This angular acceleration is constant for each particle of the body but linear acceleration is different for different particles.

Let the body contain particles of masses $m_1,\ m_2,\ m_3,\dots$ having individual linear velocities $\mathbf{v_1,\ v_2,\ v_3},\dots$ respectively. Let them each separate the axis of rotation by distances $\mathbf{r_1,\ r_2,\ r_3},\dots$

Consider the first particle. Its linear velocity is given by $v_1=r_1\omega$ where $\mathbf {\omega}$ is the angular velocity.

The linear acceleration of the first particle is given by,

$a_1=\dfrac {dv_1}{dt}\\\\a_1=\dfrac {d(r_1\omega)}{dt}\\\\a_1=r_1\cdot\dfrac {d\omega}{dt}\\\\a_1=r_1\alpha$

where $\mathbf {\alpha}$ is the angular acceleration.

The force on the first particle is given by,

$F_1=m_1a_1\\\\F_1=m_1r_1\alpha$

The torque acting on the first particle is given by,

$\tau_1=F_1r_1\\\\\tau_1=m_1(r_1)^2\alpha$

Similarly, torque acting on the second particle is,

$\tau_2=m_2(r_2)^2\alpha$

Well, torque acting on the i-th particle is,

$\tau_i=m_i(r_i)^2\alpha$

Now the net torque acting on the body is,

$\displaystyle\tau_{net}=\sum\tau\\\\\tau_{net}=m_1(r_1)^2\alpha+m_2(r_2)^2\alpha+m_3(r_3)^2\alpha+\dots\\\\\tau_{net}=\sum(mr^2\alpha)\\\\\tau_{net}=\alpha\sum(mr^2)$

But $\displaystyle\sum(mr^2)=I,$ i.e., moment of inertia.

$\therefore\ \tau=I\alpha$

Similarly, linear momentum of the first particle is given by,

$p_1=m_1v_1\\\\p_1=m_1r_1\omega$

So the angular momentum of the first particle is given by,

$L_1=p_1r_1\\\\L_1=m_1(r_1)^2\omega$

Similarly, angular momentum of the i-th particle is,

$L_i=m_i(r_i)^2\omega$

Now the net torque acting on the body is,

$\displaystyle L_{net}=\sum L\\\\L_{net}=m_1(r_1)^2\omega+m_2(r_2)^2\omega+m_3(r_3)^2\omega+\dots\\\\L_{net}=\sum(mr^2\omega)\\\\L_{net}=\omega\sum(mr^2)$

Since $\displaystyle\sum(mr^2)=I,$

$L=I\omega$

Now, come to the proof!

Consider the expression for angular momentum.

$L=I\omega$

Differentiate both sides wrt time. So,

$\dfrac {dL}{dt}=\dfrac {d(I\omega)}{dt}\\\\\dfrac {dL}{dt}=I\cdot\dfrac {d\omega}{dt}\\\\\dfrac {dL}{dt}=I\alpha\\\\\large\boxed{\dfrac {dL}{dt}=\tau}$

So we get that torque acting on a body is the rate of change of its angular momentum.

Hence the Proof!

Answer 2

Explanation:

See the proof in the attachment.