Physics Problem

Question

You measure a watch's hour and minute hands to be 7.2 mm and 11.7 mm long, respectively. In one day, by how much does the distance traveled by the tip of the minute hand exceed the distance traveled by the tip of the hour hand?

Answer 1

In one day, the hour's hand rotates twice around the watch. The tip of the hour hand travels in a circle with radius 7.2 mm, so each hour, it travels 45.239 mm (the circumference of the circle). In a full day, it travels 90.478 mm.

In one day, the minute's hand rotates 24 times around the watch. The tip of the minute hand travels in a circle with radius 11.7 mm, so each rotation it travels 73.51 mm (circumference). In a full day, it travels 1764.24 mm.

The difference is then 1673.76 mm, or 1.674 meters.

Answer 2

Just compute the distance traveled by the tip of the minute hand minus the distance traveled by the tip of the hour hand.

The minute hand goes through 24 rotations in a day ($48\pi$ radians), while the hour hand travels just 2 rotations in a day ($4\pi$ radians).

So with the equation $2\pi r$ for circumference of a circle of radius r, we get $48\pi(11.7)-4\pi(7.2)=1673.84$ mm. Slightly more than the answer given by @user2825632... so rounding errors?