star rotates once every second, (a) what is the speed or the star\'s equator and

Question

star rotates once every second, (a) what is the speed or the star's equator and (b) what is the magnitude of the particle's centripetal acceleration? (c) If the neutron star rotates faster, do the answers to (a) and (b) increase, decrease, or remain the same? What is the magnitude of the acceleration of a sprinter running at 10 m/s when rounding a turn of radius 25 m? At t 1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is (6.00 m/s 2)i + (4.00 m/s 2)j. It moves at constant speed. At time t 2 = 5.00 s, the particle's acceleration is (4.00m/s 2)i + (-6.00m/s 2)j. What is the radius of the path taken by the particle if t 2 - t 1, is less than one period? A particle moves horizontally in uniform circular motion, over a horizontal xy plane. At one instant, it moves -5.00i m/s and an acceleration of + 12.5j m/s 2.

Explanation / Answer

If the particle goes in circles, at a constant speed, then the acceleration is of constantmagnitude, and always towards the center of the circle. I suspect that's the missing bit of the puzzle.

At t1, the acceleration points in the +x direction (+6 m/s^2) i, and in the +y direction (+4 m/s^2)j. That means the particle must be in the 3rd quadrant somewhere (in fact, at angle arctan(-4/-6) ~ 213.69 deg).
At t2 (3 seconds later), the acceleration is still in the +x direction, but now in the -y direction. As 3 seconds is less than a period of the movement, the particle is now in the 2nd quadrant (at arctan(-4/6) ~ 146.31 deg).

Now, the magnitude of centripetal acceleration is given by v^2/r. The _linear_ acceleration's magnitude here is always sqrt(4^2 + 6^2) ~ 7.21 m/s^2.
(i) v^2 = r*7.21 m/s^2.

We also have that in 3 seconds, the particle has traveled a distance of s = r*theta = r*(360 deg - 213.69 + 146.31 deg) = r*5.017 radians, so that the linear speed is s/3sec = r*5.017 rad / (3 sec), or
(ii) v^2 = r^2*(5.017 rad / 3s)^2 = 2.7967 /sec^2 * r^2.

Setting (i) equal to (ii) gives
r*7.21 m/s^2 = r^2 * 2.7967 /s^2

==> r = 2.58 m.

or r = (7.21 rad/s^2)/(2.7967rad^2/s^2) ~ 2.58 meters please dont forget to rate my answer :)

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