Respuesta :
Answer:
Part a)
[tex]a_t = 0.423 m/s^2[/tex]
Part b)
[tex]a_c = 2113 m/s^2[/tex]
Part c)
[tex]d = 80 m[/tex]
Explanation:
Part a)
as we know that angular acceleration of the wheel is given as
[tex]\alpha = 13.2 rad/s^2[/tex]
now the radius of the wheel is given as
R = 3.21 cm
so the tangential acceleration is given as
[tex]a_t = R\alpha[/tex]
[tex]a_t = (0.0321)(13.2)[/tex]
[tex]a_t = 0.423 m/s^2[/tex]
Part b)
frequency of the wheel at maximum speed is given as
[tex]f = 2450 rev/min[/tex]
[tex]f = \frac{2450}{60} = 40.8 rev/s[/tex]
now we know that
[tex]\omega = 2\pi f = 2\pi(40.8) = 256.56 rad/s[/tex]
now radial acceleration is given as
[tex]a_c = \omega^2 r[/tex]
[tex]a_c = (256.56)^2(0.0321) = 2113 m/s^2[/tex]
Part c)
total angular displacement of the point on rim is given as
[tex]\Delta \theta = \omega_0 t + \frac{1}{2}\alpha t^2[/tex]
here we know that
[tex]\omega = \omega_0 + \alpha t[/tex]
[tex]256.56 = 0 + 13.2 t[/tex]
[tex]t = 19.4 s[/tex]
now angular displacement will be
[tex]\Delta \theta = 0 + \frac{1}{2}(13.2)(19.4)^2[/tex]
[tex]\Delta \theta = 2493.3 rad[/tex]
now the distance moved by the point on the rim is given as
[tex]d = R\theta[/tex]
[tex]d = (0.0321)(2493.3)[/tex]
[tex]d = 80 m[/tex]
