Question

The wheel of fortune is 2.6 meters in diameter. A contestant gives the wheel an initial velocity of 2 m/s. After rotating 540 degrees, the wheel comes to a stop. Calculate the angular acceleration of the wheel. Show your work.

Answer 1

Answer:

The angular acceleration of the wheel of fortune is -0.125 radians per square second.

Explanation:

Let suppose that wheel of fortune is decelerated at constant rate, given that wheel of fortune stops after rotating 540 degrees with an initial tangential velocity of 2 meters per second, the initial angular velocity and the kinematic expression of final angular speed as a function of angular acceleration and position are, respectively:

$\omega_{o}=\frac{v_{o}}{R}$

$\omega^{2} = \omega_{o}^{2} + 2\cdot \alpha \cdot (\theta-\theta_{o})$

Where:

$v_{o}$ - Initial tangential velocity, measured in meters per second.

$R$ - Radius of the wheel of fortune, measured in meters.

$\omega_{o}$ - Initial angular velocity, measured in radians per second.

$\omega$ - Final angular velocity, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$\theta$ - Final angular position, measured in radians.

$\theta_{o}$ - Initial angular position, measured in radians.

Angular acceleration is now cleared in the second expression:

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2 \cdot (\theta-\theta_{o})}$

Given that $v_{o} = 2\,\frac{m}{s}$ and $R = 1.3\,m$, the initial angular velocity is:

$\omega_{o} = \frac{2\,\frac{m}{s} }{1.3\,m}$

$\omega_{o} = 1.538\,\frac{rad}{s}$

Now, if $\omega = 0\,\frac{rad}{s}$, $\theta_{o} = 0\,rad$ and $\theta \approx 9.425\,rad$ (180° = π rad), the angular acceleration of the wheel is:

$\alpha = \frac{\left(0\,\frac{rad}{s} \right)^{2}-\left(1.538\,\frac{rad}{s} \right)^{2}}{2\cdot (9.425\,rad-0\,rad)}$

$\alpha = -0.125\,\frac{rad}{s^{2}}$

The angular acceleration of the wheel of fortune is -0.125 radians per square second.