Answer:
(a) [tex]\alpha=-111.26rad/s[/tex]
(b) [tex]s=4450.6in[/tex]
(c) [tex]8.66in[/tex]
Explanation:
First change the units of the velocity, using these equivalents [tex]1rev=2\pi rad[/tex] and [tex]1 min =60s[/tex]
[tex]4250rpm(\frac{2\pi rad}{1rev})(\frac{1 min}{60 s} )=445.06rad/s[/tex]
The angular acceleration [tex]\alpha[/tex] the time rate of change of the angular speed [tex]\omega[/tex] according to:
[tex]\alpha=\frac{\Delta \omega}{\Delta t}[/tex]
[tex]\Delta \omega=\omega_i-\omega_f[/tex]
Where [tex]\omega_i[/tex] is the original velocity, in the case the velocity before starting the deceleration, and [tex]\omega_f[/tex] is the final velocity, equal to zero because it has stopped.
[tex]\alpha=\frac{\Delta \omega}{\Delta t} =\frac{\omega_i-\omega_f}{4}\frac{0-445.06}{4} =\frac{-445.06}{4} =-111.26rad/s[/tex]
b) To find the distance traveled in radians use the formula:
[tex]\theta = \omega_i t + \frac{1}{2} \alpha t^2[/tex]
[tex]\theta = 445.06 (4) + \frac{1}{2}(-111.26) (4)^2=1780.24-890.12=890.12rad[/tex]
To change this result to inches, solve the angular displacement [tex]\theta[/tex] for the distance traveled [tex]s[/tex] ([tex]r[/tex] is the radius).
[tex]\theta=\frac{s}{r} \\s=\theta r[/tex]
[tex]s=890.12(5)=4450.6in[/tex]
c) The displacement is the difference between the original position and the final. But in every complete rotation of the rim, the point returns to its original position. so is needed to know how many rotations did the point in the 890.16 rad of distant traveled:
[tex]\frac{890.12}{2\pi}=141.6667[/tex]
The real difference is in the 0.6667 (or 2/3) of the rotation. To find the distance between these positions imagine a triangle formed with the center of the blade (point C), the initial position (point A) and the final position (point B). The angle [tex]\gamma=\frac{2\pi}{3}=\frac{360^o}{3}=120[/tex] is between the two sides known. Using the theorem of the cosine we can find the missing side of the the triangle(which is also the net displacement):
[tex]c^2=a^2+b^2-2abcos(\gamma)[/tex]
[tex]c^2=5^2+5^2-2(5)(5)cos(\frac{2\pi}{3} )\\c^2=25+25+25\\c^2=75\\c=5\sqrt{3}=8.66in[/tex]
