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Digiron [165]
3 years ago
12

Three solid, uniform, cylindrical flywheels, each of mass 65.0 kg and radius 1.47 m, rotate independently around a common axis t

hrough their centers. Two of the flywheels rotate in one direction at 8.94 rad/s, but the other one rotates in the opposite direction at 3.42 rad/s.
Required:
Calculate the magnitude of the net angular momentum of the system.
Physics
1 answer:
Mkey [24]3 years ago
3 0

Answer:

the angular momentum is 1015.52 kg m²/s

Explanation:

given data

mass of each flywheel, m = 65 kg

radius of flywheel, r = 1.47 m

ω1 = 8.94 rad/s

ω2 = - 3.42 rad/s

to find out

magnitude of the net angular momentum

solution

we get here Moment of inertia that is express as

I = 0.5 m r²    .................1

put here value and we get

I = 0.5 × 65 × 1.47 × 1.47

I = 70.23 kg m²

and

now we get here Angular momentum that is express as

L = I × ω    ...........................2

and Net angular momentum will be

L = 2 × I x ω1 - I × ω2

put here value and we get

L = 2 × 70.23 × 8.94 - 70.23 × 3.42

L = 1015.52 kg m²/s

so

the angular momentum is 1015.52 kg m²/s

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Breanna is standing beside a merry-go-round pushing 19° from the tangential direction and is able to accelerate the ride and her
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To solve the problem it is necessary to apply the concepts given in the kinematic equations of angular motion that include force, acceleration and work.

Torque in a body is defined as,

\tau_l = F*d

And in angular movement like

\tau_a = I*\alpha

Where,

F= Force

d= Distance

I = Inertia

\alpha = Acceleration Angular

PART A) For the given case we have the torque we have it in component mode, so the component in the X axis is the net for the calculation.

\tau= F*cos(19)*d

On the other hand we have the speed data expressed in RPM, as well

\omega_f = 10rpm = 10\frac{1rev}{1min}(\frac{1min}{60s})(\frac{2\pi rad}{1rev})

\omega_f = 1.0471rad/s

Acceleration can be calculated by

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\alpha = 0.11rad/s^2

In the case of Inertia we know that it is equivalent to

I = \frac{1}{2}mr^2 = \frac{1}{2}(750)*(2.3)^2

I = 1983.75kg.m^2

Matching the two types of torque we have to,

\tau_l=\tau_a

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Fcos(19)*2.3=1983.75(0.11)

F=100.34N

PART B) The work performed would be calculated from the relationship between angular velocity and moment of inertia, that is,

W = \frac{1}{2}I\omega_f^2

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To solve this problem it is necessary to apply the concepts related to the kinematic equations of movement description, which determine the velocity, such as the displacement of a particle as a function of time, that is to say

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v=3*10^8m/s

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Replacing we have that,

x=v*t

x=(3*10^8)(1.33)

x = 399'000.000m

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