Recall the formula,
∆<em>θ</em> = <em>ω</em>₀ <em>t</em> + 1/2 <em>α</em> <em>t</em> ²
where ∆<em>θ</em> = angular displacement, <em>ω</em>₀ = initial angular speed (which is zero because the disk starts at rest), <em>α</em> = angular acceleration, and <em>t</em> = time. Solve for the acceleration with the given information:
50 rad = 1/2 <em>α</em> (5 s)²
<em>α</em> = (100 rad) / (25 s²)
<em>α</em> = 4 rad/s²
Now find the angular speed <em>ω </em>after 3 s using the formula,
<em>ω</em> = <em>ω</em>₀ + <em>α</em> <em>t</em>
<em>ω</em> = (4 rad/s²) (3 s)
<em>ω</em> = 12 rad/s
Answer:
The magnitude of the spring force on the disk at the moment it is released is 13.8 N
Explanation:
Given;
Mass of the disk, m = 1.12 kg
spring constant, k = 230 N/m
extension produced in the spring, x = 6.0 cm
According to Hook's law, the force applied to an elastic material is directly proportional to the extension produced.
F = kx
Where;
K is the spring constant
x is the extension of the spring
F = 230 x 0.06
F = 13.8 N
Therefore, the magnitude of the spring force on the disk at the moment it is released is 13.8 N
Aaaaaaaaaaaaaaaaaaaaaaaaa