A catapult with a radial arm 3.81 m long accelerates a ball of mass 18.2 kg through a quarter circle. The ball leaves the apparatus at 49.8

Question

A catapult with a radial arm 3.81 m long accelerates a ball of mass 18.2 kg through a quarter circle. The ball leaves the apparatus at 49.8 m/s. The mass of the arm is 22.6 kg and the acceleration is constant. Hint: Use the time-independent rotational kinematics equation to find the angular acceleration, rather than the angular velocity equation.

(a) Find the angular acceleration.

rad/s2

(b) Find the moment of inertia of the arm and ball.

kg · m2

(c) Find the net torque exerted on the ball and arm.

N · m

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Delwyn 6 months 2021-07-24T02:50:14+00:00 1 Answers 10 views 0

Answers ( )

    0
    2021-07-24T02:51:31+00:00

    Answer:

    (a)\alpha = 53.73 m/s^2

    (b)   I =428 kgm^2

    (c)\tau = 428 \times 53.73  = 22996 .44Nm

    Explanation:

    GIVEN

    mass = 18.2 kg

    radial arm length = 3.81 m

    velocity = 49.8 m/s

    mass of arm = 22.6 kg

    we know using relation between linear velocity and angular velocity

    \omega = \frac{v}{l}

    \omega = \frac{49.8}{3.81} \\\omega = 12.99 rad/s

    for  angular acceleration, use the following equation.

    \omega _{f}^2 = \omega_{i}^2+2\alpha\theta

    since \omega _{i} = 0

    here  for one circle is 2 π radians.   therefore for one quarter of a circle is π/2 radians

    so   for one quarter \theta = \frac{\pi }{2}

    (12.99)^2 = 2\alpha(\frac{\pi }{2})

    on solving

    \alpha = \frac{168.74}{\pi }\\\alpha = 53.73 m/s^2

    (b)

    For the catapult,

    moment of inertia

    I = \frac{1}{2}MR^2

    I = \frac{1}{2} \times 22.6\times 3.81 \times 3.81\I = 164kg m^2

    For the ball,

    I = MR^2

    I = 18.2 \times 14.51

    I = 264 kgm^2

    so total moment of inertia =  428 kgm^2

    (c)

    \tau = I\alpha

    \tau = 428 \times 53.73  = 22996 .44Nm

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