Question

A building superintendent twirls a set of keys in a circle at the end of a cord. If the keys have a centripetal acceleration of 127 m/s^12 and the cord has a length of 0.22 m, what is the tangential speed of the keys?

Answer 1

A building superintendent twirls a set of keys in a circle at the end of a cord. If the keys have a centripetal acceleration of $127\mathrm{m/s}^2$ and the cord has a length of $0.22\mathrm{m}$, what is the tangential speed of the keys?

Solution.

$$a = 127 \frac{m}{s^2}, \quad r = 0.22m;$$ $$v - ?$$

The centripetal acceleration is:

$$a = \frac{v^2}{r},$$

$v$ – the tangential speed of the keys;

$r$ – the length of the cord.

The tangential speed of the keys:

$$v^2 = ar;$$ $$v = \sqrt{ar}.$$ $$v = \sqrt{127 \frac{m}{s^2} \cdot 0.22m} = 5.3 \frac{m}{s}.$$

Answer: The tangential speed of the keys is $v = 5.3 \frac{m}{s}$.