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GaryK [48]
3 years ago
6

• (-/1 Points) DETAILS OSCOLPHYS2016 9.5.WA.040.

Physics
1 answer:
Tanya [424]3 years ago
6 0

The force that must be exerted on the outside wheel to lift the anchor at constant speed is 6.925 x 10⁵ N.

<h3>Force exerted outside the wheel</h3>

The force exerted on the outside of the wheel can be determined by applying the principle of conservation of angular momentum as shown below.

∑τ = 0

  • Let the distance traveled by the load = 1.5 m
  • Let the radius of the wheel or position of the force = 0.45 m

∑τ = R(mg) - r(F)

rF = R(mg)

0.45F = 1.5(21,200 x 9.8)

F = 6.925 x 10⁵ N.

Thus, the force that must be exerted on the outside wheel to lift the anchor at constant speed is 6.925 x 10⁵ N.

Learn more about angular momentum here: brainly.com/question/7538238

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In a playground, there is a small merry-go-round of radius 1.20 m and mass 160 kg. Its radius of gyration is 91.0 cm. (Radius of
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Answer:

a) 145.6kgm^2

b) 158.4kg-m^2/s

c) 0.76rads/s

Explanation:

Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation 

(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and

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a) From I = MK^2

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I = 145.6kgm^2

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L = 158.4kg-m^2/s

c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:

L = Iw

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