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miskamm [114]
3 years ago
15

A heavy flywheel rotating on its central axis is slowing down because of friction in its bearings. At the end of the first minut

e of slowing, its angular speed is 0.70 of its initial angular speed of 125 rev/min. Assuming constant frictional forces, find its angular speed at the end of the second minute.
Physics
1 answer:
jonny [76]3 years ago
6 0

Answer:

ωf' =5.16 rad/s

Explanation:

Given that

Initial angular speed ,ωi= 125 rev/min    ( 1 rev/min = 0.10472 rad/s)

ωi=13.08 rad/s

The angular speed after 1 min ,ωf= 0.7 x 125 = 87.5 rev/min

ωf=9.1 rad/s

Lets angular acceleration =α

We know that

ωf = ωi + α t

Now by putting the values in the above equation we get

9.1 = 13.08 + α x 60             ( after 1 min ,t= 60 s)

\alpha =\dfrac{9.1-13.08}{60}

α = -0.066 rad/s²

The angular speed after 2 min

ωf' = ωi + α t

ωf' =13.08 - 0.066 x 2 x 60 rad/s

ωf' =5.16 rad/s

Therefore the angular speed after 2 min will be 5.16 rad/s.

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If he completes the 200 m dash in 23 s and runs at constant speed throughout the race, what is his centripetal acceleration as h
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Incomplete question.The complete one is here

A runner taking part in the 200m dash must run around the end of a track that has a circular arc with a radius curvature of 30m. The runner starts the race at a constant speed. If she completes the 200m dash in 23s and runs at constant speed throughout the race, what is her centripetal acceleration as she runs the curved portion of the track?

Answer:

a_c=2.523m/s^2

Explanation:

Given data

L_{length}=200m\\r_{radius}=30m\\t_{time}=23s

Required

Centripetal acceleration

Solution

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V_{velocity}=\frac{L_{length}}{t_{time}}\\V_{velocity}=\frac{200m}{23s}\\V_{velocity}=8.7m/s

The centripetal acceleration is given by:

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[3 points] Question: Consider a pendulum made from a uniform, solid rod of mass M and length L attached to a hoop of mass M and
aliina [53]

Answer:

I=\frac{4}{3}ML^2+2MR^2+2MRL

Explanation:

We are given that

Mass of rod=M

Length of rod=L

Mass of hoop=M

Radius of hoop=R

We have to find the moment of inertia I of the pendulum about pivot depicted at the left end of the slid rod.

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Moment of inertia of the pendulum about the pivot left end,I=\frac{1+3+12}{12}ML^2+2MR^2+2MLR

Moment of inertia  of the pendulum about the pivot left end,I=\frac{16}{12}ML^2+2MR^2+2MRL

Moment of inertia of the pendulum about the pivot left end,I=\frac{4}{3}ML^2+2MR^2+2MRL

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