Question

The peripheral speed of the tooth of a 10-in.-diameter circular saw blade is 170 ft/s when the power to the saw is turned off. The speed of the tooth decreases at a constant rate, and the blade comes to rest in 11 s. Determine the time at which the total acceleration of the tooth is 130 ft/s2.

Answer 1

Answer:

10.5 s      

Explanation:

Initial tangential speed, $u_t=170 ft/s$

Final tangential speed, $v_t=0$

Time in which it comes to stop, $t=11 s$

Diameter, d = 10 in = 0.83 ft

so, radius r = 0.415 ft

Tangential acceleration,

$a_t=\frac{v_t-u_t}{t}\\= \frac{0-170ft/s}{11s}\\=-15.45ft/s^2$

Total acceleration of the tooth is $a=130 ft/s^2$

$a^2=a_t^2+a_n^2\\a_n=\sqrt{a^2-a_t^2}\\a_n=\sqrt{130^2-15.45^2}\\a_n=120.1 ft/s^2$

$a_n=\frac{v^2}{r}\\v^2=a_n\times r\\v=\sqrt{129.1\times 0.415 ft}=7.32 ft/s$

so,

$t=\frac{v-u}{a_t} \\t=\frac{7.32-170}{-15.45} = 10.5 s$