Answer:
e). [tex]a = 0.066 m/s^2[/tex]
Explanation:
As we know that wheel is turned by 90 degree angle
so the angular speed of the wheel is given as
[tex]\omega_f^2 - \omega_i^2 = 2\alpha \theta[/tex]
now we have
[tex]\omega_f^2 - 0 = 2(0.01)(\frac{\pi}{2})[/tex]
[tex]\omega = 0.177 rad/s[/tex]
now the centripetal acceleration of the point P is given as
[tex]a_c = \omega^2 R[/tex]
[tex]a_c = (0.177)^2(2)[/tex]
[tex]a_c = 0.063 m/s^2[/tex]
tangential acceleration is given as
[tex]a_t = R\alpha[/tex]
[tex]a_t = 2(0.01)[/tex]
[tex]a_t = 0.02 m/s^2[/tex]
now net acceleration is given as
[tex]a = \sqrt{a_t^2 + a_c^2}[/tex]
[tex]a = \sqrt{0.02^2 + 0.063^2}[/tex]
[tex]a = 0.066 m/s^2[/tex]
