The drum shown in the figure has a radius of 0.40 m and a moment of inertia of 2.3 kg • m2 about an axis perpendicular to it through its center. The frictional torque at the drum axle is 3.0 N .m. A 14-m length of rope is wound around the rim. The drum is initially at rest. A constant force is applied to the free end of the rope until the rope is completely unwound and slips off. At that instant, the angular speed of the drum is 23 rad/s. The drum then slows down at a constant rate and comes to a halt. In this situation, what was the magnitude of the constant force applied to the rope?

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onyebuchinnaji onyebuchinnaji · Answer 1 · 2022-03-08T11:17:10+01:00

The constant force applied to the rope is 7.72 N.

The net torque on the drum is calculated as follows;

[tex]Fr - F_fr = I \alpha[/tex]

Linear speed, v = ωr = 23 x 0.4 = 9.2 m/s

distance traveled = 14 m

The acceleration of the drum is calculated as follows;

v² = u² + 2as

v² = 0 + 2as

v² = 2as

a = v²/2s

a = (9.2)/2(14)

a = 0.329 m/s²

[tex]\alpha = \frac{a}{r} = \frac{0.329}{0.4} = 0.82 \ rad/s^2[/tex]

The constant force applied to the rope is calculated as follows;

[tex]Fr - F_fr = I \alpha\\\\Fr = I \alpha + F_fr\\\\F = \frac{ I \alpha + F_fr}{r} \\\\F = \frac{2.3(0.82) \ + \ 3(0.4)}{0.4} \\\\F = 7.72 \ N[/tex]

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