Answer:
The kinetic energy of the disk is 254.4 J.
Explanation:
The kinetic energy of the disk is given by:
[tex] K = \frac{1}{2}I \omega^{2} [/tex]
The moment of inertia of the solid disk is:
[tex] I = \frac{1}{2}mr^{2} [/tex]
The mass is:
[tex] m = \frac{P}{g} [/tex]
[tex]I = \frac{Pr^{2}}{2g} = \frac{803 N*(1.41 m)^{2}}{2*9.81 m/s^{2}} = 81.4 kgm^{2}[/tex]
Now, we need to find the angular acceleration as follows:
[tex] \tau = I \alpha [/tex]
Also, the torque is related to the tangential force:
[tex] \tau = F\times r [/tex]
[tex] F\times r = I \alpha [/tex]
[tex] \alpha = \frac{F \times r}{I} = \frac{49.7 N*1.41 m}{81.4 kg*m^{2}} = 0.86 rad/s^{2} [/tex]
Now, we can find the angular speed:
[tex] \omega_{f} = \omega_{i} + \alpha t [/tex]
[tex] \omega_{i} = 0 [/tex] since it is started from rest
[tex] \omega_{f} = 0.86 rad\s^{2}*2.90 s = 2.50 rad/s [/tex]
Finally, the kinetic energy of the disk is:
[tex] K = \frac{1}{2}I \omega^{2} = \frac{1}{2}81.4 kgm^{2}*(2.50 rad/s)^{2} = 254.4 J [/tex]
Therefore, the kinetic energy of the disk is 254.4 J.
I hope it helps you!
