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

Answer 1

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)

$\tau = I\alpha$

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