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

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

(b) I =428
`kgm^2`

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

Step-by-step 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 = (v)/(l)`

`\omega = (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 = (\pi )/(2)`

`(12.99)^2 = 2\alpha((\pi )/(2))`

on solving

`\alpha = (168.74)/(\pi )\\\alpha = 53.73 m/s^2`

(b)

For the catapult,

moment of inertia

`I = (1)/(2)MR^2`

`I = (1)/(2) * 22.6* 3.81 * 3.81\I = 164kg m^2`

For the ball,

`I = MR^2`

`I = 18.2 * 14.51`

`I = 264 kgm^2`

so total moment of inertia = 428
`kgm^2`

(c)

`\tau = I\alpha`

`\tau = 428 * 53.73 = 22996 .44Nm`