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I am teaching myself basic mechanics from a standing start. I am trying to understand Angular Acceleration and have set myself a problem to solve. My answer 'feels' wrong, so I'd like some help to understand if I've misunderstood, or miscalculated anything. I've taken many liberties with rounding, please ignore, this is more about the basic process/theory than accuracy. Thanks in advance!!

Problem

An object is travelling around a circle with a radius of 40m. It's speed at (A) is calculated as 50mph. 5 seconds later, it's speed at (B) is calculated as 40mph. Determine the Angular Acceleration.

Basic conversions

Circumference = $2\pi r = 251\text{ m}$

Velocity (A) = $22\text{ m/s}$

Velocity (B) = $18\text{ m/s}$

Angular Velocity at (A)

251 / 22 = 11.4. Therefore one full revolution would take 11.4 seconds.

$$\omega = \theta/t$$

$$\omega = 2\pi /t$$

$$\omega = 2\pi/11.4$$

$$\omega = 0.55 \text{ rad/s}$$

Angular Velocity at (B)

$251/18 = 13.9$. Therefore one full revolution would take $13.9$ seconds.

$$\omega = \theta/t$$

$$\omega = 2\pi /t$$

$$\omega = 2\pi/13.9$$

$$\omega = 0.45\text{ rad/s}$$

Angular Acceleration

$$\alpha = \frac{d\omega}{dt} $$

$$\alpha = \frac{0.45 - 0.55}{5 - 0}$$

$$\alpha = -0.1 / 5$$

Answer to Problem

$\alpha = -0.02\text{ }\mathrm{rad/s^2}$

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  • $\begingroup$ It looks correct to me. Why does that feel wrong to you? $\endgroup$
    – Lelesquiz
    Commented Jun 15, 2014 at 10:40
  • $\begingroup$ If it looks correct, that's really great news. I thought the number seemed too small, but this wasn't based on anything whatsoever. It probably came from a lack of comprehension of rad/s2. Thank you for confirming. $\endgroup$
    – user2781522
    Commented Jun 15, 2014 at 10:43
  • $\begingroup$ One thing to note is that this is correct if you would assume that the angular acceleration is constant. $\endgroup$
    – fibonatic
    Commented Jun 15, 2014 at 10:48
  • $\begingroup$ Of course, that angular acceleration you calculated implies that the acceleration (or rather deceleration) is uniform, i.e. constant over the path from (A) to (B). If the object traveled without deceleration for quite a while from (A) to (C) (where (C) is somewhere between (A) and (B)) and was decelerated only then, the deceleration would be more abrupt and $\alpha$ would be more negative. $\endgroup$
    – Jonas Greitemann
    Commented Jun 15, 2014 at 10:49
  • $\begingroup$ As a preliminary comment, where does m stand for meters, and when does it stand for miles? $\endgroup$
    – DJohnM
    Commented Jun 15, 2014 at 17:18

1 Answer 1

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The results are correct, but the process can be simplified:

One of the benefits of using radians is that the angular quantity (distance, velocity, or acceleration) is the linear quantity divided by the radius.

Once you find $$Velocity (A) = 22 m/s$$just divide by the radius, $40\text{ m}$, and you have the angular velocity. You can do the same for the velocity at B.

Even simpler, the linear velocity between A and B drops from 22 m/s to 18 m/s in 5 seconds, for an average linear acceleration of $-0.8\text{ m/s}^2.$ Again, just divide by 40 to get the angular acceleration...

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