Answer:
(A) Angular speed 40 rad/sec
Rotation = 50 rad
(b) 37812.5 J
Explanation:
We have given moment of inertia of the wheel [tex]I=25kgm^2[/tex]
Initial angular velocity of the wheel [tex]\omega _0=10rad/sec[/tex]
Angular acceleration [tex]\alpha =15rad/sec^2[/tex]
(a) We know that [tex]\omega =\omega _0+\alpha t[/tex]
We have given t = 2 sec
So [tex]\omega =10+15\times 2=40rad/sec[/tex]
Now [tex]\Theta =\omega _0t+\frac{1}{2}\alpha t^2=10\times 2+\frac{1}{2}\times 15\times 2^2=50rad[/tex]
(b) After 3 sec [tex]\omega =10+15\times 3=55rad/sec[/tex]
We know that kinetic energy is given by [tex]Ke=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 25\times 55^2=37812.5J[/tex]
