Respuesta :
Answer:
(a)[tex]\alpha = 53.73 m/s^2[/tex]
(b) I =428 [tex]kgm^2[/tex]
(c)[tex]\tau = 428 \times 53.73 = 22996 .44Nm[/tex]
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
[tex]\omega = \frac{v}{l}[/tex]
[tex]\omega = \frac{49.8}{3.81} \\\omega = 12.99 rad/s[/tex]
for angular acceleration, use the following equation.
[tex]\omega _{f}^2 = \omega_{i}^2+2\alpha\theta[/tex]
since [tex]\omega _{i} = 0[/tex]
here for one circle is 2 π radians. therefore for one quarter of a circle is π/2 radians
so for one quarter [tex]\theta = \frac{\pi }{2}[/tex]
[tex](12.99)^2 = 2\alpha(\frac{\pi }{2})[/tex]
on solving
[tex]\alpha = \frac{168.74}{\pi }\\\alpha = 53.73 m/s^2[/tex]
(b)
For the catapult,
moment of inertia
[tex]I = \frac{1}{2}MR^2[/tex]
[tex]I = \frac{1}{2} \times 22.6\times 3.81 \times 3.81\I = 164kg m^2[/tex]
For the ball,
[tex]I = MR^2[/tex]
[tex]I = 18.2 \times 14.51[/tex]
[tex]I = 264 kgm^2[/tex]
so total moment of inertia = 428 [tex]kgm^2[/tex]
(c)
[tex]\tau = I\alpha[/tex]
[tex]\tau = 428 \times 53.73 = 22996 .44Nm[/tex]
