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Vedmedyk [2.9K]

2 years ago

15

man stands on a platform that is rotating (without friction) with an angular speed of 1.2 rev/s; his arms are outstretched and h

e holds a brick in each hand.The rotational inertia of the system consisting of the man, bricks, and platform about the central vertical axis of the platform is 6.0 k g times m squared. If by moving the bricks the man decreases the rotational inertia of the system to 2.0 k g times m squared, what is the resulting angular speed of the platform in rad/s? Express to 3 sig figs.

1 answer:

-Dominant- [34]2 years ago

3 0

Answer:

w₂ = 22.6 rad/s

Explanation:

This exercise the system is formed by platform, man and bricks; For this system, when the bricks are released, the forces are internal, so the kinetic moment is conserved.

Let's write the moment two moments

initial instant. Before releasing bricks

L₀ = I₁ w₁

final moment. After releasing the bricks

$L_{f}$ = I₂W₂

L₀ = L_{f}

I₁ w₁ = I₂ w₂

w₂ = I₁ / I₂ w₁

let's reduce the data to the SI system

w₁ = 1.2 rev / s (2π rad / 1rev) = 7.54 rad / s

let's calculate

w₂ = 6.0/2.0 7.54

w₂ = 22.6 rad/s

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2 years ago

Read 2 more answers

A 45kg skater and a 60kg skater are standing still, holding hands on frictionless ice. They push away from each other in opposit

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Answer:

The 60 kg skater is traveling at 2.63 m/s in the opposite direction from the 45 kg skater.

Explanation:

The velocity of the 60 kg skater can be found by conservation of linear momentum:

$P_{i} = P_{f}$

$m_{a}v_{i_{a}} + m_{b}v_{i_{b}} = m_{a}v_{f_{a}} + m_{b}v_{f_{b}}$ (1)

Where:

$m_{a}$ : is the mass of the first skater = 45 kg

$m_{b}$ : is the mass of the second skater = 60 kg

$v_{i_{a}}$ : is the initial speed of the first skater = 0 (he is standing still)

$v_{i_{b}}$ : is the initial speed of the second skater = 0 (he is standing still)

$v_{f_{a}}$ : is the final speed of the first skater = 3.5 m/s

$v_{f_{b}}$ : is the final speed of the second skater =?

By replacing the above values into equation (1) and solving for $v_{f_{b}}$ we have:

$0 = m_{a}v_{f_{a}} + m_{b}v_{f_{b}}$

$v_{f_{b}} = \frac{-m_{a}v_{f_{a}}}{m_{b}} = \frac{-45 kg*3.5 m/s}{60 kg} = -2.63 m/s$

The minus sign is because the 60 kg skater is moving in the opposite direction from the other skater.

Therefore, the 60 kg skater is traveling at 2.63 m/s in the opposite direction from the 45 kg skater.

I hope it helps you!

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