A particle of mass m moves due to the force =-\alpha m\vec{r} where alpha>0 is a const, and \vec{r} is the radius vector of the particle with respect to the origin.
Initially....\vec{r}=r_0\hat{i} & \vec{v}=v_0\hat{j} where \vec{v} denotes the velocity vector....
Find the trajectory of motion of the particle.
49accn=\alpha \vec{r}
or, {\color{red} \frac{\partial v}{\partial t}}=\alpha \vec{r}
or, {\color{blue} v\frac{\partial v}{\partial r}}=\alpha \vec{r}
or,v{\partial v}=-\alpha r{\partial r}{\color{green} }
or,\int v{\partial v}=\int -\alpha r{\partial r} \Rightarrow v^2=-\alpha r^2
is this approach correct??????
shall i proceed with this??
66A few things about the motion:
1) The force is always directed towards the origin.
2) The angular momentum about the origin does not change. (Why?)
3) For all times, the velocity vector is perpendicular to the position vector. (How?)
4) The distance of the particle remains constant from the origin.
5) The acceleration vector is of the form -\omega^2\vec{r}.
The deduction: The particle moves in a circle of radius r0 uniformly with a speed v0.
1sir
2) force passes through the origin thus no net torque. (α \propto -\vec{r})
