A uniform-density 7 kg disk of radius 0.20 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is wrapped tightly around the disk, and you pull on the string with a constant force of 42 N through a distance of 0.9 m. Now what is the angular speed

Answer:

The angular speed is 23.24 rad/s.

Explanation:

Given;

mass of the disk, m = 7 kg

radius of the disk, r = 0.2 m

applied force, F = 42 N

distance moved by disk, d = 0.9 m

The torque experienced by the disk is calculated as follows;

τ = F x d = I x α

where;

I is the moment of inertia of the disk = ¹/₂mr²

α is the angular acceleration

F x r = ¹/₂mr² x α

The angular acceleration is calculated as;

[tex]\alpha = \frac{2Fr}{mr^2} \\\\\ \alpha = \frac{2F}{mr}\\\\\alpha = \frac{2 \times 42 }{7 \times 0.2} \\\\\alpha = 60 \ rad/s^2[/tex]

The angular speed is determined by applying the following kinematic equation;

[tex]\omega _f^2 = \omega_i ^2 + 2\alpha \theta[/tex]

initial angular speed, ωi = 0

angular distance, θ = d/r = 0.9/0.2 = 4.5 rad

[tex]\omega _f^2 = 2\alpha \theta\\\\\omega _f = \sqrt{2\alpha \theta} \\\\\omega _f = \sqrt{2 \times 60 \times 4.5} \\\\\omega _f = 23.24 \ rad/s[/tex]

Therefore, the angular speed is 23.24 rad/s.

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