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3 years ago

7

At the end of cylindrical rod of length l = 1 m and mass M = 1 kg rotating horizontaly along the vertical axis in its center wit

h an angular speed  a small mouse of mass m = 0.02 kg is located Determine the angular speed 
of this double system when a mouse will move to the center of rod close to the rotating axis, taking into
account that the mouse can be treated as the material point.

1 answer:

matrenka [14]3 years ago

3 0

Answer:

w = 0.943 rad / s

Explanation:

For this problem we can use the law of conservation of angular momentum

Starting point. With the mouse in the center

L₀ = I w₀

Where The moment of inertia (I) of a rod that rotates at one end is

I = 1/3 M L²

Final point. When the mouse is at the end of the rod

$L_{f}$ = I w + m L² w

As the system is formed by the rod and the mouse, the forces during the movement are internal, therefore the angular momentum is conserved

L₀ = L_{f}

I w₀ = (I + m L²) w

w = I / I + m L²) w₀

We substitute the moment of inertia

w = 1/3 M L² / (1/3 M + m) L² w₀

w = 1 / 3M / (M / 3 + m) w₀

We substitute the values

w = 1/3 / (1/3 + 0.02) w₀

w = 0.943 w₀

To finish the calculation the initial angular velocity value is needed, if we assume that this value is w₀ = 1 rad / s

w = 0.943 rad / s

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also we know that

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now we have

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3 years ago

An electrical motor spins at a constant 2662.0 rpm. If the armature radius is 6.725 cm, what is the acceleration of the edge of

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Answer:

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3 years ago

Two charges q1 and q2, that are distance d apart , repel each other with a force of 6.40 N. what would be the force between two

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Q: Two charges q1 and q2, that are distance d apart , repel each other with a force of 6.40 N. what would be the force between two charges q1'=2q1 and q2'=3q2 that that are distance d apart?

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The force = 38.4 N

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From coulombs law,

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When, F = 6.4 N, r = d m.

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When q₁' = 2q₁, q₂' = 3q₂, r = d cm

F = k(2q₁)(3q₂)/d²

F = 6kq₁q₂/d².......................... Equation 2

Dividing Equation 1 by equation 2

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4 years ago

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3 years ago

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anzhelika [568]

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Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

$KE_{initial} + PE_{initial} = KE_{final} + PE_{final}$

where

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Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

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7 0

3 years ago

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