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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.

Developer
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2 Answers2

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To calculate the average acceleration over the $30$-min period, we just need the initial velocity $\mathbf{u}$ (at 4:00), the final velocity $\mathbf{v}$, (at 4:30), and the time interval $t$ ($30$ minutes or $1800$ seconds).

Since the hand moves with a uniform speed, the magnitude of velocity would be the same for both $\mathbf{u}$ and $\mathbf{v}$ ($r \omega$, radius times the magnitude of angular velocity). Only the directions will be different. Assuming the conventional orientation of a wall clock, it's towards right (horizontal) for $u$ and towards left for $v$.

Basically, the magnitude of acceleration $a$ would be

$$a = \frac{2 \cdot r \cdot \omega}{t}$$

Now, the tip of the minute's hand covers $2 \pi$ radians every hour (or $3600$ seconds). So,

$$\omega = \frac{2 \pi}{3600}$$

Do the substitution and calculation and you should be good.

Haris
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The hint from Intelligenti pauca is perfect: regard the situation at $t_2$ and what changed since $t_1$.
As you showed already, $v_1$ is about 2,5 cm/min (to the right), after 30 min $v_2$ is -2,5 cm/min (to the left and for this $\lt 0$). Thus $\Delta v$ is 5 cm/min, therefore the acceleration $$\displaystyle a=\frac{-5 cm/min}{30 min}=-\frac{1}{6}\frac{cm}{min^2}$$(roughly).

m-stgt
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