In a model generator, a 310-turn rectangular coil 0.075 m by 0.18 m rotates with an angular frequency of 12.2 rad/s in a uniform
1 answer:
This question involves the concepts of the induced emf, magnetic field, and angular frequency.
The maximum emf induced in the coil is "32.68 volts".
The following formula can be used to find out the induced emf in the coil.
But, for maximum induced emf sin function should have a value of 1. Therefore,
where,
E = induced emf = ?
N = No. of turns in the coil = 310 turns
ω = angular frequency = 12.2 rad/s
A = Area of coil = (0.075 m)(0.18 m) = 0.0135 m²
B = magnetic field = 0.64 T
Therefore,
<u>E = 32.68 volts</u>
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Answer:
A) = 1.44 kg m², B) moment of inertia must increase
Explanation:
The moment of inertia is defined by
I = ∫ r² dm
For figures with symmetry it is tabulated, in the case of a cylinder the moment of inertia with respect to a vertical axis is
I = ½ m R²
A very useful theorem is the parallel axis theorem that states that the moment of inertia with respect to another axis parallel to the center of mass is
I = + m D²
Let's apply these equations to our case
The moment of inertia is a scalar quantity, so we can add the moment of inertia of the body and both arms
=
+ 2
= ½ M R²
The total mass is 64 kg, 1/8 corresponds to the arms and the rest to the body
M = 7/8 m total
M = 7/8 64
M = 56 kg
The mass of the arms is
m’= 1/8 m total
m’= 1/8 64
m’= 8 kg
As it has two arms the mass of each arm is half
m = ½ m ’
m = 4 kg
The arms are very thin, we will approximate them as a particle
= M D²
Let's write the equation
= ½ M R² + 2 (m D²)
Let's calculate
= ½ 56 0.20² + 2 4 0.20²
= 1.12 + 0.32
= 1.44 kg m²
b) if you separate the arms from the body, the distance D increases quadratically, so the moment of inertia must increase