Respuesta :
Answer:
(A) 2.4 N-m
(B) [tex]0.035kgm^2[/tex]
(C) 315.426 rad/sec
(D) 1741.13 J
(E) 725.481 rad
Explanation:
We have given mass of the disk m = 4.9 kg
Radius r = 0.12 m, that is distance = 0.12 m
Force F = 20 N
(a) Torque is equal to product of force and distance
So torque [tex]\tau =Fr[/tex], here F is force and r is distance
So [tex]\tau =20\times 0.12=2.4Nm[/tex]
(B) Moment of inertia is equal to [tex]I=\frac{1}{2}mr^2[/tex]
So [tex]I=\frac{1}{2}\times 4.9\times 0.12^2=0.035kgm^2[/tex]
Torque is equal to [tex]\tau =I\alpha[/tex]
So angular acceleration [tex]\alpha =\frac{\tau }{I}=\frac{2.4}{0.035}=68.571rad/sec^2[/tex]
(C) As the disk starts from rest
So initial angular speed [tex]\omega _{0}=0rad/sec[/tex]
Time t = 4.6 sec
From first equation of motion we know that [tex]\omega =\omega _0+\alpha t[/tex]
So [tex]\omega =0+68.571\times 4.6=315.426rad/sec[/tex]
(D) Kinetic energy is equal to [tex]KE=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 0.035\times 315.426^2=1741.13J[/tex]
(E) From second equation of motion
[tex]\Theta =\omega _0t+\frac{1}{2}\alpha t^2=0\times 4.6+\frac{1}{2}\times 68.571\times 4.6^2=725.481rad[/tex]
