Respuesta :
Solution :
Given :
Diameter, D = 30 m
∴ Radius, R = 15 m
Angular acceleration, α = 0.044 [tex]$rad/s^2$[/tex]
a). Velocity of the rider just as he completes a quarter of a turn is :
[tex]$\omega_i = 0 \ rad/s$[/tex]
[tex]$\theta = \frac{\pi}{2}\ rad$[/tex]
V = R ω
∴[tex]$\omega^2_f=\omega^2_i + 2 \alpha \theta$[/tex]
[tex]$\omega^2_f=0+ 2 \times 0.04 \times \frac{\pi}{2}$[/tex]
[tex]$\omega^2_f=0.1256$[/tex]
[tex]$\omega = \sqrt{0.1256}$[/tex]
[tex]$=0.354 \ rad/s$[/tex]
∴ [tex]$V=R \omega_f$[/tex]
[tex]$= 15 \times 0.354$[/tex]
= 5.32 m/s up
b). Tangential acceleration
[tex]$a_T= \alpha R$[/tex]
= 0.04 x 15
[tex]$= 0.600 \ m/s^2$[/tex] up
Radial acceleration,
[tex]$a_r=\frac{V^2}{R}$[/tex]
[tex]$=\frac{(5.32)^2}{15}$[/tex]
[tex]$= 1.88 \ m/s^2$[/tex] towards center.
c). Final angular velocity
given : [tex]$V_0 = 6 \ m/s$[/tex]
[tex]$\omega_i = 0.354 \ rad/s$[/tex]
[tex]$\alpha = 0.0400 \ rad/s^2$[/tex]
[tex]$\omega_f = \frac{6}{15} \ rad/s$[/tex]
[tex]$\omega^2_f=\omega^2_i + 2 \alpha \theta$[/tex]
[tex]$\left(\frac{6}{15}\right)^2=(0.354)^2 + 2 \times 0.04 \times \theta$[/tex]
[tex]$\theta = \frac{0.16-0.1256}{2 \times 0.04}$[/tex]
= 0.43 rad
or [tex]$\theta = 0.43 \times \frac{180}{\pi}$[/tex]
[tex]$24.6^\circ$[/tex]
