Question about circular acceleration

[Extemporary]

I have a question regarding acceleration in circular motion that I was hoping someone on this forum might be able to help with.

I had a physics test in which I did unusually poorly, and I am unsure why. The teacher posted the solution to the question online, but it left me unclear as to what his intent was. The question was as follows:

A traditional watch hand has a second hand 1.45 cm long from centre to tip. What is the magnitude of the average acceleration between 30 s and 50 s?

In class, we had only briefly covered circular motion, but I remembered the equation we'd used:

Since the motion of the hand is constant the acceleration should be constant as well. To be safe, I tried the equation twice, once over the 20 s interval and 1/3 the circumference, and once over the whole 60 s. Both times I got an answer of 1.6*10^-4 m/s^2. However, when I received the marked test back, I had been given 1.5/8 with the note "not reasonable". I know I lost 1/2 a mark on my diagram; I'll admit it was a lousy one. The other mark I did get was for my final answer, since, while different, it complied with notation and significant digits. Several days later he posted his solution on his website:

What was wrong with my method? Every circular question we'd done previously had worked with the equation I used. I considered that perhaps my solution was for instantaneous acceleration rather than average, but for an object with constant speed shouldn't those two be identical? Can anyone see where I went wrong?

[N_las]

Your Formula calculates the magnitude of the instantaneous acceleration, but it ignores the direction of it completely. You are right, that instantaneous and average acceleration of an object with constant velocity shoud be identical, but the watch hand doesn't have a constant velocity, since it changes direction all the time.

For example: the average acceleration of the clock hand during 60 seconds is zero. Because after 60 seconds, the clock hand has exactly the same velocity (magnitude and direction) as at the beginning, so in 60 seconds there was no change in velocity. That means zero average acceleration.

In your Problem: You have to calculate the velocity of the clock hand at 30 seconds as vector, because the direction is important. Then you have to calculate the velocity at 50 seconds as vector. Now you substract the 50s-vector from the 30s-vector, to get the vector of the change in velocity. If you divide this vector by 20 seconds you get the vector of the average acceleration. Since the magnitude is wanted, you then have to calculate the magnitude of this vector.

[Extemporary]

Ah, I understand where the confusion came from. In the lesson on circular motion we never distinguished between average and instantaneous acceleration. We only did three problems where we found the circular acceleration, and they all used the formula I tried. He even specifically said that for these questions we would only need the magnitude. When I saw the question, I saw circular motion and a magnitude-only response, and assumed it was the stuff we did in class.

Thanks for the help.

[Death by Reentry]

You could always print off your teachers solution mark the mistakes and hand it back to him/her with a 3 out of 8. Not recommended, but the errors are pretty bad.

[Sillychris]

Oh man, that's a burn. I could see myself making that same mistake.

[Doozler]

The question itself is a bit sneaky as a minor re-wording completely changes the meaning of the question. If it had said:

"What is the average magnitude of the acceleration"

Rather than

"What is the magnitude of the average acceleration "

Then your answer would have been correct. I don't like maths questions that rely on English comprehension - especially when the "obvious" misreading is a more "sensible" question. Still, now you know your teacher asks sneaky questions you can look out for them.

[Death by Reentry]

Which one?

The lack of a consistent coordinate system. Might be nit-picking as the teacher does add the values,but if you're teaching students you should be able to do this.

The mysterious way the teacher has sin(30)=.78722. I can't even figure out how this error would occur.

The way they are off an order of magnitude for the change in x velocity. They realize it at the end, but did not correct any of the previous steps to account for this.