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svet-max [94.6K]
3 years ago
6

Determine the total moment of inertia of a merry-go round with 5 children sitting on it. Of the five children, four are seated a

t the edge of the merry-go-round, and the fifth child is sitting exactly at the center. The children have the same mass of 30.0 kg. You may consider the merry-go-round as a cylinder of mass 150 kg and a radius 2.0 meters.
Physics
1 answer:
BaLLatris [955]3 years ago
7 0

Answer:

Explanation:

Given that,

We have five children.

Each of mass m =30kg

They sit on a merry go round

Mass of Merry go round M= 150kg

Radius of Merry go round is r =2m

Four children sit at the edge of the merry go round but one child sit at the centre.

The four child that sit at the edge are 2m from the centre of the merry go round but the one at the centre is 0m from the centre

Moment of inertia?

Moment of inertia is given as

I=Σmi•ri²

For the question, the moment of inertia is the combination of inertial of child and the merry go round

I= I(merry go round) + I(four child)+ I(last child)

The merry go round is assumed to be a solid cylinder, so it is going to have the moment of inertia of solid cylinder

Then,

I(merry go round ) =½ Mr²

Also, Four of the child has the same moment of inertia, they are 2m form the centre of the merry go round why the last child has no moment of inertia

I= I(merry go round) + I(four child) +I(last child )

I= ½Mr² + 4mr² + mr'²

I = ½ × 150 ×2² + 4×30×2² + 30×0²

I = 300 +480+0

I = 780 kgm²

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Answer:

The magnitude of force per unit length of one wire on the other is 2.7945\times 10^{-5}\ N and the direction is away from one another

The magnitude of force per unit length of one wire on the other is 2.7945\times 10^{-5}\ N and the direction is towards each other.

Explanation:

\mu_0 = Vacuum permeability = 4\pi\times 10^{-7}\ N/A^2

i_1 = Current in first wire = 2.9 A

i_2 = Current in second wire = 5.3 A

r = Gap between the wires = 11 cm

Force per unit length

F_{12}=F_{21}=\frac{\mu_0 i_1i_2}{2\pi r}\\ =\frac{4\pi\times 10^{-7}\times 2.9\times 5.3}{2\pi 0.11}\\ =2.7945\times 10^{-5}\ N

The magnitude of force per unit length of one wire on the other is 2.7945\times 10^{-5}\ N and the direction is away from one another

F_{12}=F_{21}=\frac{\mu_0 i_1i_2}{2\pi r}\\ =\frac{4\pi\times 10^{-7}\times 2.9\times 5.3}{2\pi 0.11}\\ =2.7945\times 10^{-5}\ N

The magnitude of force per unit length of one wire on the other is 2.7945\times 10^{-5}\ N and the direction is towards each other.

7 0
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