Answer:
a) 79.7rad/s
b) -18.7rad/s^2
c) 53m
Explanation:
We will use the MKS system of unit, so:
[tex]v=89.0km/h=89.0\frac{km}{h}*\frac{1000m.h}{3600km.s}=24.7m/s\\\\d=62.0cm=62.0cm*\frac{0.01m}{1cm}=0.62m[/tex]
now, The angular speed is given by:
[tex]\omega=\dfrac{v}{\frac{d}{2}}\\\\\\\omega=\frac{24.7m/s}{0.31m}=79.7rad/s[/tex]
in order to obtain the angular acceleration we have to apply the following formula:
[tex](\omega_f)^2=(\omega_o)^2+2\alpha*\theta\\\\\alpha=-\frac{(\omega_o)^2}{2*rev*2\pi}\\\\\alpha=-\frac{(79.7m/s)^2}{2*27*2\pi}=-18.7rad/s^2[/tex]
The linear displacement is given by:
[tex]d_l=\theta*r\\d_l=rev*2\pi*\frac{d}{2}\\\\d_l=27*2\pi*0.31m=53m[/tex]
