Question

A circular saw spins at 6000 rpm , and its electronic brake is supposed to stop it in less than 2 s. As a quality-control specialist, you're testing saws with a device that counts the number of blade revolutions. A particular saw turns 75 revolutions while stopping.Does it meet its specs?
a.yes

b.no

Answer 1

Answer:

a. yes

Explanation:

The initial speed of the circular saw is:

$\dot n_{o} = \frac{6000}{60} \,\frac{rev}{s}$

$\dot n_{o} = 100\,\frac{rev}{s}$

Deceleration rate needed to stop the circular saw is:

$\ddot n = -\frac{100\,\frac{rev}{s} }{2\,s}$

$\ddot n = - 50\,\frac{rev}{s^{2}}$

The number of turns associated with such deceleration rate is:

$\Delta n = \frac{\dot n^{2}}{2\cdot \ddot n}$

$\Delta n = \frac{\left(100\,\frac{rev}{s} \right)^{2}}{2\cdot \left(50\,\frac{rev}{s^{2}} \right)}$

$\Delta n = 100\,rev$

Since the measured number of revolutions is lesser than calculated number of revolution, the circular saw meets specifications.