Respuesta :
Answer:
[tex]\omega_f=5.097\ rad.s^{-1}[/tex]
Explanation:
Given:
- mass of uniform disk, [tex]m=38.7\ kg[/tex]
- radius of the disk, [tex]r=0.24\ m[/tex]
- initial velocity of the disk, [tex]\omega_i=0\ rad,s^{-1}[/tex]
- magnitude of the force acted upon the rim of the disk, [tex]F=30\ N[/tex]
- angular displacement of the disk, [tex]\theta=0.32\times 2\pi=2.011\ rad[/tex]
Moment of inertia of the given disk:
[tex]I=\frac{1}{2}\times m.r^2[/tex]
[tex]I=0.5\times 38.7\times 0.24^2[/tex]
[tex]I=1.11456\ kg.m^2[/tex]
Also the torque due to the force is given as:
[tex]\tau=F\times r[/tex]
[tex]\tau=30\times 0.24[/tex]
[tex]\tau=7.2\ N.m[/tex]
Torque in terms of angular acceleration:
[tex]\tau=I.\alpha[/tex]
[tex]\alpha=\frac{\tau}{I}[/tex]
where:
[tex]\alpha=[/tex] angular acceleration
[tex]\alpha=\frac{7.2}{1.11456}[/tex]
[tex]\alpha=6.4599\ rad.s^{-2}[/tex]
Now using the equation of motion:
[tex]\omega_f^2=\omega_i^2+2\times \alpha.\theta[/tex]
[tex]\omega_f^2=0+2\times 6.4599\times 2.011[/tex]
[tex]\omega_f=5.097\ rad.s^{-1}[/tex]
