Calculating the magnitude of the average acceleration of a clock hand

Question

I've been stuck on the following question from Isaac Physics for quite some time now and I'm not really sure where to even begin:

The time shown on a clock changes from 4:00 to 4:30. The minute hand, of length 25 cm, moves smoothly halfway around the face. The movement of the tip of the minute hand can be thought of as lots of small displacement vectors taking the tip from the old to the new position. Where vector answers are required, give them in terms of unit vectors $i$ and $j$ pointing from 12 o'clock to 6 o'clock and from 9 o'clock to 3 o'clock respectively. A convenient unit of speed will be $cm min^{-1}$ (centimetres per minute).

Calculate the magnitude of the average acceleration of the minute hand during the half hour.

My initial thought was to calculate the average velocity of the minute hand during this time period. The average speed of the tip is 2.6 cm/minute. The horizontal components of velocity cancel in the average over half hour. The vertical component is the speed times $sinθ$ where $θ$ is the angle of the hand to the vertical. I averaged this over $π$ radians to obtain an average velocity of 1.7 cm/minute. To calculate the average acceleration, I thought I could divide the average velocity by a further 30 minutes to obtain $0.057 cmmin^{-2}$. However, this answer is incorrect, but I do not know why.

Answer 1

My initial thought was to calculate the average velocity of the minute hand during this time period.

This isn't the correct approach. We're interested in the acceleration, meaning how the velocity changes. If we simply look at the average velocity, then we're throwing away any information about how the velocity changed during the time period under consideration.

At the top of the hour (4:00) the tip of the minute hand is moving in the +x (positive horizontal) direction at 2.6 cm/min with no vertical component. At the bottom of the hour it is moving in the -x direction with no vertical component.

So the total change in velocity of the tip between those times is -5.2 cm/min. Since it took 30 minutes to make this change, the average acceleration over that period is

$$\bar{a}=\frac{-5.2\ {\rm cm/min}}{30\ {\rm min}}$$

This is about $-0.17\ {\rm cm/min^2}$. Since we've only been asked for the magnitude, that would be $0.17\ {\rm cm/min^2}$

The key point is that to calculate the average acceleration, we only need the initial and final velocities, and the time period over which the change occurs.

Answer 2

The average acceleration is:

$$ \langle\vec a\rangle = \frac{\vec v_f - \vec v_i}{\Delta t} $$

Since:

$$ \vec v_i = \vec v_f = 0 $$

it is zero.

Now if you just consider the average tip speed of $\bar v = 2.6\,{\rm cm/min}$, then you don't need to consider the clock face in real space. In velocity space, the clock hand tip traces out a semi-circle with length:

$$ \int_{\theta=0}^{\theta=\pi} \bar vd\theta = \pi\bar v $$

So the average acceleration is:

$$ \bar a = \frac{\pi\bar v}{\Delta t} $$

Take your pick.