Answer:
The time is [tex]t_r =21.58 \ s[/tex]
Explanation:
From the question we are told that
The diameter of the circular saw is [tex]d = 10 \ in = \frac{10}{12} = 0.833 \ feet[/tex]
The peripheral speed is [tex]u = 230 \ ft/s[/tex]
The time taken for the blade to come to rest is t = 17 s
The total acceleration of the tooth considered is [tex]a = 130 \ ft/s^2[/tex]
Generally the radius of the blade is mathematically represented as
[tex]r = \frac{0.833}{2}= 0. 4165 \ feet[/tex]
Generally the tangential acceleration of the blade is mathematically represented as
[tex]a__{t}} = \frac{v - u}{t}[/tex]
Here v is the final velocity of the tooth of the blade which is zero since the blade came to rest
so
[tex]a__{t}} = \frac{0 - 230}{ 17}[/tex]
=> [tex]a__{t}} = - 13.53 \ ft/s^2[/tex]
Generally the total acceleration of the tooth of the blade is mathematically represented as
[tex]a = \sqrt{a_t^2 + a_r^2}[/tex]
Here [tex]a_r[/tex] is the radial acceleration , now making [tex]a_r[/tex] the subject of the formula we have that
[tex]130= \sqrt{13.56 ^2 + a_r^2}[/tex]
=> [tex]a_r = \sqrt{130^2 -(- 13.56)^2}[/tex]
=> [tex]a_r = 129.3 \ m/s^2[/tex]
Generally radial acceleration is mathematically represented as
[tex]a_r = \frac{v_r^2}{r}[/tex]
Here [tex]v_r[/tex] is the velocity at which the total acceleration is 130 ft/s2.
=> [tex]v_r = \sqrt{a_r * r }[/tex]
=> [tex]v_r = \sqrt{129.3 * 0.4165 }[/tex]
=> [tex]v_r = 7.34 \ m/s[/tex]
Generally the time at which the total acceleration is 130 ft/s2. is mathematically represented as
[tex]t_r = \frac{7.34 - 300}{a_t}[/tex]
=> [tex]t_r = \frac{7.34 - 300}{-13.56}[/tex]
=> [tex]t_r =21.58 \ s[/tex]
