Answer :
Answer:
part a)
[tex]\omega = 3.7 rad/s[/tex]
Part b)
v = 44.4 m/s
Part c)
[tex]a_t = 8.88 m/s^2[/tex]
Part d)
[tex]a_r = 164.3 m/s^2[/tex]
Explanation:
As we know that the angle at any instant of time is given as
[tex]\theta = 0.370 t^2[/tex]
part a)
as we know that rate of change in angle with time is angular speed
so here we have
[tex]\omega = \frac{d\theta}{dt}[/tex]
[tex]\omega = \frac{d}{dt}(0.370 t^2)[/tex]
[tex]\omega = 0.74 t[/tex]
at t= 5s we will have
[tex]\omega = (0.74)(5) = 3.7 rad/s[/tex]
Part b)
as we know the relation between linear speed and angular speed is given as
[tex]v = R\omega[/tex]
[tex]v = (12.0)(3.7) [/tex]
v = 44.4 m/s
Part c)
now in order to find angular acceleration we know that
[tex]\alpha = \frac{d\omega}{dt}[/tex]
[tex]\alpha = \frac{d(0.74 t)}{dt} [/tex]
[tex]\alpha = 0.74 rad/s^2[/tex]
now tangential acceleration is given as
[tex]a_t = R\alpha[/tex]
[tex]a_t = 12.0(0.74) [/tex]
[tex]a_t = 8.88 m/s^2[/tex]
Part d)
radial acceleration is given as
[tex]a_r = \frac{v^2}{R}[/tex]
[tex]a_r = \frac{44.4^2}{12}[/tex]
[tex]a_r = 164.3 m/s^2[/tex]