Answer:
0.76 rad/s^2
Explanation:
First, we convert the original and final velocity from rev/s to rad/s:
[tex]v_o = 3.3\frac{rev}{s} * \frac{2\pi rad}{1rev} =20.73 rad/s[/tex]
[tex]v_f = 6.4\frac{rev}{s} * \frac{2\pi rad}{1rev}=40.21 rad/s[/tex]
Now, we need to find the number of rads that the tire rotates in the 250m path. We use the arc length formula:
[tex]D = x*r \\x = \frac{D}{r} = \frac{250m}{0.64m/2} = 781.25 rads[/tex]
Now, we just use the formula:
[tex]w_f^2-w_o^2=2\alpha*x[/tex]
[tex]\alpha =\frac{w_f^2-w_o^2}{2x} = \frac{(40.21rad/s)^2-(20.73rad/s)^2}{2*781.25rad} = 0.76 rad/s^2[/tex]
Explanation:
The given data is as follows.
Initial velocity, u = [tex]3.3 rev/s \times 2 \pi rad/rev[/tex]
= 20.724 rad/sec
Final velocity, v = [tex]6.4 rev/s \times 2 \times pi rad/rev[/tex]
= 40.192 rad/sec
Now, we will calculate the value of angular rotation (d) as follows.
d = No. of revolutions × [tex]2 \pi rad/rev[/tex]
d = [tex](\frac{250 m}{2 \pi r}) \times 2 \pi[/tex]
= [tex]\frac{250}{0.32}[/tex]
= 781.25 rad
Also, we know that,
[tex]v^{2} = u^{2} + 2ad[/tex]
or, a = [tex]\frac{(v^{2} - u^{2})}{2d}[/tex]
= [tex]\frac{(40.192)^{2} - (20.724)^{2})}{2 \times 781.25}[/tex]
= [tex]\frac{1615.39 - 429.48}{1562.5}[/tex]
= 0.7589 [tex]rad/s^{2}[/tex]
Thus, we can conclude that the tire's angular acceleration is 0.7589 [tex]rad/s^{2}[/tex].
