My Math Forum Circular Acceleration

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 September 20th, 2013, 09:21 AM #1 Member   Joined: Aug 2013 Posts: 69 Thanks: 0 Circular Acceleration Hi, question 1:" a steel ball that is accelerated on a circular track of radius 72cm reaches a speed of 4.0m/s over three circuits, starting from rest. How long after the ball starts to move does its radial accel = tangential accel. How many turns does that ball take for that to happen? Assume contant tang accel. 2.An object moves in a cirtial circle of radius 4m, the speed of the object is 12m/s when the object is 30deg above the lowest point along the circle, and 8m/s when the object is at the top of the circle. the tan accel is uniform. what is the total accel of the objet when its speed is 8m/s I have ways of doing this myself. but im not getting the correct answers all the way through, I think my intuition about accell are off. ..f I can accurately figure out a tan i can do some wokr. Zen~
September 25th, 2013, 09:35 PM   #2
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Re: Circular Acceleration

Hi Zen,

Here is how I worked the problems. I hope that my answers are correct (I have my doubts about problem #2 though).

Units are: time, s; distance, m; speed, m/s; acceleration, m/s^2

Problem #1

First find the acceleration a:

$v=at, \qquad x=\frac{1}{2}at^2$

$v=4, \qquad x=(0.72)(2\pi)(3)=13.572 \$ (the 3 is 3 revolutions)

$4=at \to t=\frac{4}{a}, \qquad 13.572=\frac{1}{2}at^2 \$ substituting and solving for a we get

$a= 0.58945$

Quote:
 How long after the ball starts to move does its radial accel = tangential accel.
$a=\frac{v^2}{r} \to 0.58945=\frac{v^2}{0.72} \to v=0.65146$

$t=\frac{v}{a}=\frac{0.65146}{0.58945} \to \fbox{t=1.1052}$

Quote:
 How many turns does that ball take for that to happen?
$x=\frac{1}{2}at^2=\frac{1}{2}(0.58945)(1.0152)^2=0 .3600$

$N_{revs}=\frac{0.36}{2\pi(0.72)} \to \fbox{N_{revs}=0.07958}$

Problem #2

I interpret the problem as the object decelerating as it moves through $\ \frac{5}{6}\pi \$ radians.

Again, let's find the tangential a

$v=v_o-at \to v=12-at$

$x=v_ot-\frac{1}{2}at^2 \to x=12t-\frac{1}{2}at^2$

At the point of interest, the velocity is 8 and the distance traveled is $\ \frac{5}{6}\pi(4)=10.472$

$8=12-at \to t=\frac{4}{a} \qquad 10.472=\frac{48}{a}-\frac{1}{2}\cancel{a}\frac{16}{a^{\cancel{2}}} \to 10.472=\frac{40}{a}$

$a=3.8197 \ \text{tangential acceleration}$

$a=\frac{v^2}{r}=\frac{8^2}{4}=16 \ \text{radial acceleration}$

The resultant acceleration magnitude is $\ \sqrt{16^2+3.8197^2}=\fbox{16.45} \$ with an angle of direction that is

$\theta=\arctan{\frac{3.81971}{16}=\fbox{13.43 \ \text{degrees}}$ away from the radius line pointing back in the direction that the object came since it is decelerating.

I hope this is close.

 Tags acceleration, circular

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