Answer:
10.5 s
Explanation:
Initial tangential speed, [tex]u_t=170 ft/s[/tex]
Final tangential speed, [tex]v_t=0[/tex]
Time in which it comes to stop, [tex]t=11 s[/tex]
Diameter, d = 10 in = 0.83 ft
so, radius r = 0.415 ft
Tangential acceleration,
[tex]a_t=\frac{v_t-u_t}{t}\\= \frac{0-170ft/s}{11s}\\=-15.45ft/s^2[/tex]
Total acceleration of the tooth is [tex]a=130 ft/s^2[/tex]
[tex]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[/tex]
[tex]a_n=\frac{v^2}{r}\\v^2=a_n\times r\\v=\sqrt{129.1\times 0.415 ft}=7.32 ft/s[/tex]
so,
[tex]t=\frac{v-u}{a_t} \\t=\frac{7.32-170}{-15.45} = 10.5 s[/tex]
