Question #105795
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
1
Expert's answer
2020-03-19T07:52:48-0400

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

(1) r(t)=cos(t2)i^+sin(t2)j^\vec{r}(t)=cos (t^2)\hat{i} +sin(t^2)\hat{j} .

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

(2) r=rr=cos2(t2)+sin2(t2)=1|\vec r|=\sqrt{\vec r\cdot \vec r}=\sqrt{cos^2(t^2)+sin^2(t^2)}=1

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

(3) r˙=2tsin(t2)i^+2tcos(t2)j^\dot {\vec{r}}=-2t sin (t^2)\hat{i} +2t cos(t^2)\hat{j} and acceleration

(4) r¨=[2sin(t2)4t2cos(t2)]i^+[2cos(t2)4t2sin(t2)]j^\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

r¨r^=[2sin(t2)4t2cos(t2)]cos(t2)+[2cos(t2)4t2sin(t2)]sin(t2)==2sin(t2)cos(t2)4t2cos2(t2)+2cos(t2)sin(t2)4t2sin2(t2)=4t2\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 4t2-4t^2 .


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

Assignment Expert
19.03.20, 17:14

Dear GRACE, You are welcome. We are glad to be helpful. If you liked our service, please press a like-button beside the answer field. Thank you!

GRACE
19.03.20, 17:04

the solution is excellent. thank you

LATEST TUTORIALS
APPROVED BY CLIENTS