At t? = 3.00 s, the acceleration of a particle moving at...

At t? = 3.00 s, the acceleration of a particle moving at constant speed in counterclockwise circular motion is $\vec{a_1} = (8.00 \text{ m/s}^2)\hat{i} + (8.00 \text{ m/s}^2)\hat{j}$ At t? = 6.00 s (less than one period later), the acceleration is $\vec{a_2} = (8.00 \text{ m/s}^2)\hat{i} - (8.00 \text{ m/s}^2)\hat{j}$ The period is more than 3.00 s. What is the radius of the circle?

Transcript

00:01 In this problem, the acceleration at two different times are given and we have to find the radius of the circle.

00:07 So, acceleration is centripetal acceleration because it is moving at constant speed, so tangential acceleration will be zero.

00:15 So, centripetal acceleration is magnitude of the acceleration at any of the two times.

00:23 So it will be x component square plus y component square equal to 8 .0 .8.

00:33 06 meter per second square.

00:35 This is the centripetal acceleration.

00:38 Now at time t equal to 1 second and time t equal to 3 seconds there are two values of acceleration.

00:46 Their magnitude is obviously same because centripetal acceleration has same magnitude throughout the motion if the speed is constant but their directions are different.

01:00 Now we can see that a1 dot product with a 2 will be equal to 1i plus 8 j times .product 8 i minus 1 j so dot product is 1 into 8 plus 8 8 into minus 1 sorry 1 into 8 we multiply x component with x component and y component with y component to find a dot product and add them and this becomes 0 so the angle between 2 acceleration is 90 degree obviously because their dot product is 0.

01:45 So in a circle, let us say the initial direction was a1, then final direction will be a2, which is at 90 degree with a1.

02:03 So clearly from initial to final position, t equal to 1, 2 t equal to 3 seconds, they cover quarter of the circle.

02:13 And the length of quarter of circle is 2 pi r, which is the circumferumphant...

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