Respuesta :
Answer:
(a) [tex]\alpha = - 1.32\ rev/m^{2}[/tex]
(b) [tex]\theta = 13674\ rev[/tex]
(c) [tex]\alpha_{tan} = 8.75\times 10^{- 4}\ m/s^{2}[/tex]
(d) [tex]a = 22.458\ m/s^{2}[/tex]
Solution:
As per the question:
Angular velocity, [tex]\omega = 190\ rev/min[/tex]
Time taken by the wheel to stop, t = 2.4 h = [tex]2.4\times 60 = 144\ min[/tex]
Distance from the axis, R = 38 cm = 0.38 m
Now,
(a) To calculate the constant angular velocity, suing Kinematic eqn for rotational motion:
[tex]\omega' = \omega + \alpha t[/tex]
[tex]omega' [/tex] = final angular velocity
[tex]\omega [/tex] = initial angular velocity
[tex]\alpha [/tex] = angular acceleration
Now,
[tex]0 = 190 + \alpha \times 144[/tex]
[tex]\alpha = - 1.32\ rev/m^{2}[/tex]
Now,
(b) The no. of revolutions is given by:
[tex]\omega'^{2} = \omega^{2} + 2\alpha \theta[/tex]
[tex]0 = 190^{2} + 2\times (- 1.32) \theta[/tex]
[tex]\theta = 13674\ rev[/tex]
(c) Tangential component does not depend on instantaneous angular velocity but depends on radius and angular acceleration:
[tex]\alpha_{tan} = 0.38\times 1.32\times \frac{2\pi}{3600} = 8.75\times 10^{- 4}\ m/s^{2}[/tex]
(d) The radial acceleration is given by:
[tex]\alpha_{R} = R\omega^{2} = 0.32(80\times \frac{2\pi}{60})^{2} = 22.45\ rad/s[/tex]
Linear acceleration is given by:
[tex]a = \sqrt{\alpha_{R}^{2} + \alpha_{tan}^{2}}[/tex]
[tex]a = \sqrt{22.45^{2} + (8.75\times 10^{- 4})^{2}} = 22.458\ m/s^{2}[/tex]
