Question

Starting from rest, a Ferris wheel of diameter 30.0 m undergoes an angular acceleration of 0.0400 rad/s2. A certain rider is at the lowest point of the wheel just as it starts to move.
A) Find the velocity of the rider just as he completes a quarter of a turn.
B) Find the radial and tangential components of his acceleration at the same point.
C) How much farther must the wheel turn before the rider attains a speed of 6.00 m/s (the maximum that occurs during the ride)?
a. 5.32 m/s, up.
b. 0.600 m/s2, up.
1.88 m/s2 toward center.
d. 24.6°.

Answer 1

Solution :

Given :

Diameter, D = 30 m

∴ Radius, R = 15 m

Angular acceleration, α = 0.044 $rad/s^2$

a). Velocity of the rider just as he completes a quarter of a turn is :

$\omega_i = 0 \ rad/s$

$\theta = \frac{\pi}{2}\ rad$

V = R ω

∴$\omega^2_f=\omega^2_i + 2 \alpha \theta$

 $\omega^2_f=0+ 2 \times 0.04 \times \frac{\pi}{2}$

$\omega^2_f=0.1256$

$\omega = \sqrt{0.1256}$

   $=0.354 \ rad/s$

∴ $V=R \omega_f$

     $= 15 \times 0.354$

    = 5.32 m/s up

b). Tangential acceleration

   $a_T= \alpha R$

        = 0.04 x 15

       $= 0.600 \ m/s^2$  up

Radial acceleration,

$a_r=\frac{V^2}{R}$

   $=\frac{(5.32)^2}{15}$

   $= 1.88 \ m/s^2$ towards center.

c). Final angular velocity

   given : $V_0 = 6 \ m/s$

              $\omega_i = 0.354 \ rad/s$

              $\alpha = 0.0400 \ rad/s^2$

$\omega_f = \frac{6}{15} \ rad/s$

$\omega^2_f=\omega^2_i + 2 \alpha \theta$

$\left(\frac{6}{15}\right)^2=(0.354)^2 + 2 \times 0.04 \times \theta$

$\theta = \frac{0.16-0.1256}{2 \times 0.04}$

  = 0.43 rad

or $\theta = 0.43 \times \frac{180}{\pi}$

      $24.6^\circ$