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 w

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.

in progress 0
Gia Bảo 5 years 2021-08-21T18:41:31+00:00 1 Answers 43 views 0

Answers ( )

  1. minhkhang
    0
    2021-08-21T18:42:48+00:00
    Reply

    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.

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )