Answer to Question #105795 in Differential Equations for Grace

Question

particle moves in a circle according to the equation \\(\\bar{r}=cos (t^2)\\hat{i} +sin(t^2)\\hat{j}\\). The magnitude of the normal component of the acceleration at time \\(t\\) is

Answer 1

Let's make sure that the particle really moves around the circle.

(1) $\vec{r}(t)=cos (t^2)\hat{i} +sin(t^2)\hat{j}$ .

To do this, we find the vector (1) module

(2) $|\vec r|=\sqrt{\vec r\cdot \vec r}=\sqrt{cos^2(t^2)+sin^2(t^2)}=1$

Thus $\vec r$ is the unit vector $\vec r=\hat r$ . The particle moves in a circle with unit radius. Find its velocity

(3) $\dot {\vec{r}}=-2t sin (t^2)\hat{i} +2t cos(t^2)\hat{j}$ and acceleration

(4) $\ddot {\vec{r}}=[-2sin (t^2)-4t^2cos(t^2)]\hat{i} +[2 cos(t^2)-4t^2sin(t^2)]\hat{j}$ .

By virtue of (2) the normal component of the acceleration is

$\ddot {\vec r}\cdot \hat r=[-2sin (t^2)-4t^2cos(t^2)]cos(t^2)+[2 cos(t^2)-4t^2sin(t^2)]sin(t^2)=\\=-2sin (t^2)cos(t^2)-4t^2cos^2(t^2)+2 cos(t^2)sin(t^2)-4t^2sin^2(t^2)=-4t^2$

A negative acceleration sign indicates that it is directed to the center of the circle along which the particle moves.

Answer: The normal component of the acceleration is $-4t^2$ .