angular momentum

adelinetohyj

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Posts: 6

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9 years ago

A drum has a radius of 0.40 m and a moment of inertia of 4.5 kg m2. The frictional
torque of the drum axle is 3.0 N m. A 15 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 velocity of the drum is 13 rad/s. The drum then decelerates and comes to
a halt. The constant force applied to the rope is closest to
(A) 7.5 N
(B) 14 N
(C) 18 N
(D) 27 N
(E) 33 N

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f_zah1

wrote...

9 years ago

E) 33N

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adelinetohyj Author

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9 years ago

hi can you show me the full solution? thanks a lot

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f_zah1

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9 years ago

Okay..here's how you work it out.

let the force be F Newtons

Work(net) = Wnet = ½I(ωₔ² -ωᵢ ²) -------------------(ωₔ = final ω and ωᵢ = initial)

Wnet = ½I(13² -0 ²) = ½ x 4.5 x 169 = 2.25 x 169 = 380.25 Joules ----------------------> (i)

the torque τ provided by force F = F(0.40)

net τ = F(0.40) - fr τ = 0.4F - 3

from (i) = Wnet = τ(net) x θ rad = [0.4F - 3] 15/[2π(0.40)]

=> 380.25 = 5.9659[0.4F - 3]

=> 0.4F = 3 + (380.25)/5.9659

=> 0.4F = 66.7041

=> F = 2.5 x 66.70411 = 166.760 N

So the answer is E

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Adeline

wrote...

9 years ago

Hi how do you get 33N as the answer? The working you provide give 166N

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f_zah1

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9 years ago

Let me check and get back to you..

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girl4God

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9 years ago

Thanks so much!

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