Answer:
Angular acceleration will be equal to [tex]119.30m/sec^2[/tex]
Explanation:
We have given that wheel tires are moving with constant rate of [tex]\omega =2.97rev/sec=2.97\times 2\pi =19.10rad/sec[/tex]
Distance is given r = 32.7 cm = 0.327 m
Linear velocity is given by [tex]v=\omega r=19.01\times 0.327=6.24m/sec[/tex]
Wee to have to find the angular acceleration
Angular acceleration is equal to [tex]a=\frac{v^2}{r}=\frac{6.24^2}{0.327}=119.30m/sec^2[/tex]
So angular acceleration will be equal to [tex]119.30m/sec^2[/tex]
