Answer:
The kinetic energy of the merry-goround after 3.62 s is 544J
Explanation:
Given :
Weight w = 745 N
Radius r = 1.45 m
Force = 56.3 N
To Find:
The kinetic energy of the merry-go round after 3.62 = ?
Solution:
Step 1: Finding the Mass of merry-go-round
[tex]m = \frac{ weight}{g}[/tex]
[tex]m = \frac{745}{9.81 }[/tex]
m = 76.02 kg
Step 2: Finding the Moment of Inertia of solid cylinder
Moment of Inertia of solid cylinder I =[tex]0.5 \times m \times r^2[/tex]
Substituting the values
Moment of Inertia of solid cylinder I
=>[tex]0.5 \times 76.02 \times (1.45)^2[/tex]
=> [tex]0.5 \times 76.02\times 2.1025[/tex]
=> [tex]79.91 kg.m^2[/tex]
Step 3: Finding the Torque applied T
Torque applied T = [tex]F \times r[/tex]
Substituting the values
T = [tex]56.3 \times 1.45[/tex]
T = 81.635 N.m
Step 4: Finding the Angular acceleration
Angular acceleration ,[tex]\alpha = \frac{Torque}{Inertia}[/tex]
Substituting the values,
[tex]\alpha = \frac{81.635}{79.91}[/tex]
[tex]\alpha = 1.021 rad/s^2[/tex]
Step 4: Finding the Final angular velocity
Final angular velocity ,[tex]\omega = \alpha \times t[/tex]
Substituting the values,
[tex]\omega = 1.021 \times 3.62[/tex]
[tex]\omega = 3.69 rad/s[/tex]
Now KE (100% rotational) after 3.62s is:
KE = [tex]0.5 \times I \times \omega^2[/tex]
KE =[tex]0.5 \times 79.91 \times 3.69^2[/tex]
KE = 544J
