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Sagot :
D. The minimum braking torque that must be applied is -11.2 Nm.
Moment of inertia of the uniform cylinder
The moment of inertia of the uniform cylinder is calculated as follows;
I = ¹/₂MR²
where;
- M is mass of the cylinder
- R is radius of the cylinder
I = (0.5)(8)(0.1²)
I = 0.04 kgm²
Minimum braking torque
τ = -Iα
where;
- α is angular acceleration
α = ω/t
α = (80,000 x 2π/rev x 1 min/60s) / (30 s)
α = (80,000 x 2π)/(60 x 30)
α = 279.25 rad/s²
τ = - ( 0.04 kgm²) x (279.25)
τ = -11.2 Nm
Thus, the minimum braking torque that must be applied is -11.2 Nm.
Learn more about minimum braking torque here: https://brainly.com/question/6845979
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