Answer:
The angular speed is 0.13 rev/s
Explanation:
From the formula
[tex]\tau = I\alpha[/tex]
Where [tex]\tau[/tex] is the torque
[tex]I[/tex] is the moment of inertia
[tex]\alpha[/tex] is the angular acceleration
But, the angular acceleration is given by
[tex]\alpha = \frac{\omega}{t}[/tex]
Where [tex]\omega[/tex] is the angular speed
and [tex]t[/tex] is time
Then, we can write that
[tex]\tau = \frac{I\omega}{t}[/tex]
Hence,
[tex]\omega = \frac{\tau t}{I}[/tex]
Now, to determine the angular speed, we first determine the Torque [tex]\tau[/tex] and the moment of inertia [tex]I[/tex].
Here, The torque is given by,
[tex]\tau = rF[/tex]
Where r is the radius
and F is the force
From the question
r = 3.00 m
F = 195 N
∴ [tex]\tau = 3.00 \times 195[/tex]
[tex]\tau = 585[/tex] Nm
For the moment of inertia,
The moment of inertia of the solid disk is given by
[tex]I = \frac{1}{2}MR^{2}[/tex]
Where M is the mass and
R is the radius
∴[tex]I = \frac{1}{2} \times 325 \times (3.00)^{2}[/tex]
[tex]I = 1462.5[/tex] kgm²
From the question, time t = 2.05 s.
Putting the values into the equation,
[tex]\omega = \frac{\tau t}{I}[/tex]
[tex]\omega = \frac{585 \times 2.05}{1462.5}[/tex]
[tex]\omega = 0.82[/tex] rad/s
Now, we will convert from rad/s to rev/s. To do that, we will divide our answer by 2π
0.82 rad/s = 0.82/2π rev/s
= 0.13 rev/s
Hence, the angular speed is 0.13 rev/s,