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olga55 [171]
3 years ago
14

A motor has a rotor (with armature) of moment of inertia ????m . The rotor is attached to a gear box of gear ratio G > 1. The

output of the gearbox is attached to a mass whose moment of inertia is ????. What will be the moment of inertia ‘felt’ by the motor? What will be the moment of inertial ‘felt’ by someone who is rotating the mass ???? by hand, to turn the motor? Which of the two is large
Physics
1 answer:
Inessa [10]3 years ago
3 0

Answer:

hello your question is incomplete attached below is the complete question

answer : The moment of inertial felt by someone ( J ) is greater that the moment of inertia felt by the motor  i.e. J > Jm

Explanation:

Gear ratio G > 1

a) Determine the moment of inertia felt by the motor

moment of inertia felt by Motor = moment of Inertia at the armature

b) Determine the moment of inertial felt by someone who is rotating the mass by hand

moment of inertia felt by someone is = J

The moment of inertial felt by someone ( J ) is greater that the moment of inertia felt by the motor

attached below is a detailed solution

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The maximum kinetic energy, maximum potential energy and the maximum mechanical energy are equal to 7.56J.

<h3>What is simple harmonic motion?</h3>

Simple harmonic motion, in physics, repetitive movement back and forth through an equilibrium, or central, position, so that the maximum displacement on one side of this position is equal to the maximum displacement on the other side.

Simple Harmonic Motion

The given equation of the simple harmonic motion is

x=3.5 sin (\frac{\pi }{2t} + \frac{5\pi }{4} )

Data;

ω = π/2

k = 1.254N/m

Solving this

\frac{dx}{dt} = -3.5 X \frac{\pi }{2} cos (\frac{x\pi t}{2}+\frac{5\pi }{4}  )

Let's calculate the maximum velocity.

V_{m} =\frac{3.5\pi }{2}

This is only possible when cos θ = -1

The maximum kinetic energy is

K_m =\frac{1}{2} mv^2 = \frac{1}{2} X \frac{500}{1000} X \frac{7^2\pi ^2}^{4} ^2

w^2 = \frac{k}{m} \\k = w^2m\\k = \frac{\pi ^2}{4} X \frac{500}{1000} \\k =1.254 N/m

Using the value of spring constant, we can find the maximum potential energy.

P.E =\frac{1}{2} k x^2\\P.E =\frac{1}{2} X 1.234 X 3.5^2 \\P.E = 7.56 J

The maximum potential energy is 7.56J

The maximum mechanical energy is equal to the sum of maximum potential energy and the maximum kinetic energy.

ME = K.E + P.E

ME = 7.56J

From the calculations above, the maximum kinetic energy, maximum potential energy and the maximum mechanical energy are equal to 7.56J.

Learn more on simple harmonic motion here;

brainly.com/question/15556430

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