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Elodia [21]
4 years ago
5

A uniform disk has a mass of 3.7 kg and a radius of 0.40 m. The disk is mounted on frictionless bearings and is used as a turnta

ble. The turntable is initially rotating at 30 rpm. A thin-walled hollow cylinder has the same mass and radius as the disk. It is released from rest, just above the turntable, and on the same vertical axis. The hollow cylinder slips on the turntable for 0.20 s until it acquires the same final angular velocity as the turntable. What is the final angular momentum of the system
Physics
1 answer:
nevsk [136]4 years ago
4 0

Answer:

1.25 kgm²/sec

Explanation:

Disk inertia, Jd =

Jd = 1/2 * 3.7 * 0.40² = 0.2960 kgm²

Disk angular speed =

ωd = 0.1047 * 30 = 3.1416 rad/sec

Hollow cylinder inertia =

Jc = 3.7 * 0.40² = 0.592 kgm²

Initial Kinetic Energy of the disk

Ekd = 1/2 * Jd * ωd²

Ekd = 0.148 * 9.87

Ekd = 1.4607 joule

Ekd = (Jc + 1/2*Jd) * ω²

Final angular speed =

ω² = Ekd/(Jc+1/2*Jd)

ω² = 1.4607/(0.592+0.148)

ω² = 1.4607/0.74

ω² = 1.974

ω = √1.974

ω = 1.405 rad/sec

Final angular momentum =

L = (Jd+Jc) * ω

L = 0.888 * 1.405

L = 1.25 kgm²/sec

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A system of pulleys is used to raise a load of bricks that weighs 1,700 newtons. The force applied to the pulley is 340 newtons.
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4 0
3 years ago
Read 2 more answers
A block of mass m=1.2 kg is held at rest against a spring with a force constant k=730N/m. Initially the spring is compressed a d
igor_vitrenko [27]

Answer:

Explanation:

potential energy of compressed spring

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Kinetic energy after crossing the rough patch

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= 3.174 J

Loss of energy

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This loss is due to negative work done by frictional force

work done by friction = friction force x width of patch

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= .44 x 1.2 x 9.8 x .05

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3 years ago
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Explanation:

Given

\mu_s =0.330

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Frictional force F_s

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\mu_sN=ma_{min}

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