Question

A boy is riding on a merry-go-round that rotates at a constant speed. If the merry-go- round makes 3.0 revolutions in 6.0 seconds. What is its angular speed in revolutions/second and in radians/second? Show all the formulas and every single step of your solution. bila boho gelbsolwetodorowse below. S m ood tolaswi 0.00 Toxocara sottol o b) The boy is pushing on the edge (2.4 m from the center) of the merry-go-round to make it rotate. If the boy increases the merry-go-round's angular speed from 0.50 rev/s to 1.5 rev/s in 2.0 s, what is the merry-go-round's angular acceleration? Show all the formulas and every single step of your solution. w ostale voxed out of nowded co e So 0.00 Go to slingssit Wo c) What is the number of rotations during the 2.0 seconds in part b? Show all the formulas and every single step of your solution.

Answer 1

To solve this problem we will apply the concepts related to the kinematic equations of angular motion, for which the angular velocity is defined as the change between the angular position in a given time, and in turn, it is possible to define the change in velocity
`( \Delta \omega = \omega_ {final} - \omega_ {initial})` depending on angular acceleration and time or angular acceleration and angular position.

PART A) The angular velocity is defined as,

`\omega = (\Delta \theta)/(\Delta t)`

`\omega = (3rev)/(6s) = 0.5rev/s ((2\pi rad)/(1\rev))`

`\omega = 3.14rad/s`

PART B) The change in angular velocity in terms of angular acceleration and time

`\omega_f- \omega_i =\alpha t`

`1.5rev/s - 0.5rev/s =\alpha (2s)`

`\alpha = 0.5rev/s^2 ((2\pi rad)/(1rev))`

`\alpha = 3.14rad/s^2`

PART C) The change in angular velocity in terms of angular acceleration and position

`\omega_f^2 - \omega_i^2=2\alpha\theta`

`1.5^2=0.5^2+2*0.5\theta`

`\theta = 2rev`