Respuesta :
Answer:
Final angular velocity, [tex]\omega_f=25\ rad/s[/tex]
Explanation:
Given that,
Torque, [tex]\tau=50\ N-m[/tex]
Moment of inertia, [tex]I=20\ kg m^2[/tex]
Time, t = 10 s
Initial angular velocity, [tex]\omega_i=0[/tex]
The relation between the torque and the moment of inertia is :
[tex]\tau=I\times \alpha[/tex]
[tex]\alpha[/tex] is the angular acceleration
[tex]\alpha =\dfrac{\tau}{I}[/tex]
[tex]\alpha =\dfrac{50}{20}[/tex]
[tex]\alpha =2.5\ rad/s^2[/tex]
Using first equation of kinematics to find the final angular velocity of the wheel. It is given by :
[tex]\omega_f=\omega_i+\alpha t[/tex]
[tex]\omega_f=0+2.5\times 10[/tex]
[tex]\omega_f=25\ rad/s[/tex]
So, the final angular velocity of the wheel is 25 rad/s. Hence, this is the required solution.
