A catapult with a radial arm 3.94 m long accelerates a ball of mass 20.9 kg thro

Question

A catapult with a radial arm 3.94 m long accelerates a ball of mass 20.9 kg through a quarter circle. The ball leaves the apparatus at 41.1 m/s. The mass of the arm is 28.7 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.

Explanation / Answer

a)initial angular velocity=0
final angular velocity=linear veloicty/radius=10.43 rad/sec

angle travelled=(1/4)*2*pi=0.5*pi

using final anglular veloicty^2-initial angular veloicty^2=2*angular acceleration*angle travelled

we get angular acceleration=34.62 rad/sec^2

b) moment of inertia of the arm and ball:

assumption: catapult is a rod rotating around its end point

then its moment of inertia is given as (1/3)*mass*length^2
=148.51 kg.m^2

moment of inertia of the ball=mass*distance ^2=324.44 kg.m^2

so net moment of inertia=472.95 kg.m^2

c) net torque=moment of inertia*angular acceleration=16373.53 N.m

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