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tino4ka555 [31]
3 years ago
9

A 40.0 kg wheel, essentially a thin hoop with radius 0.810 m, is rotating at 438 rev/min. It must be brought to a stop in 21.0 s

.
A) How much work must be done to stop it?
B) What is the required average power?
Physics
2 answers:
garik1379 [7]3 years ago
4 0

Answer:

(a) Workdone = -27601.9J

(b) Average required power = 1314.4W

Explanation:

Mass of hoop,m =40kg

Radius of hoop, r=0.810m

Initial angular velocity Winitial=438rev/min

Wfinal=0

t= 21.0s

Rotation inertia of the hoop around its central axis I= mr²

I= 40 ×0.810²

I=26.24kg.m²

The change in kinetic energy =K. E final - K. E initail

Change in K. E =1/2I(Wfinal² -Winitial²)

Change in K. E = 1/2 ×26.24[0-(438×2π/60)²]

Change in K. E= -27601.9J

(a) Change in Kinetic energy = Workdone

W= 27601.9J( since work is done on hook)

(b) average required power = W/t

=27601.9/21 =1314.4W

Ilia_Sergeevich [38]3 years ago
3 0

Answer:

Explanation:

Given that,

Mass of wheel M = 40kg

Radius of thin hoop R= 0.81m

Angular speed ωi = 438 rev/mins

1rev = 2πrad

ωi = 438 × 2π/60 rad/sec

ωi = 45.87 rad/s

Time take to stop t = 21s

Moment of inertial of a hoop around it central axis can be calculated using

I = MR²

I = 40 × 0.81²

I = 26.24 kgm²

Then,

Change in kinetic energy is given as

∆K.E = Kf — Ki

∆K.E = ½Iωf² — ½Iωi²

Then final angular velocity is zero, since the wheel comes to rest

∆K.E = ½Iωf² — ½Iωi²

∆K.E = ½I × 0² — ½ × 26.24 × 45.87²

∆K.E = 0 — 27,609.43

∆K.E = —27,609.43 J

B. Power

Power = Workdone/Time take

The change in kinetic energy of the loop is equal to the net work done on it

∆K•E = W = —27,609.43

Power = |W| / t

P = 27,609.43/21

P = 1314.73 W

The average power to stop the loop is 1314.73 watts

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\texttt{ }

<h3>Further explanation</h3>

Let's recall Elastic Potential Energy formula as follows:

\boxed{E_p = \frac{1}{2}k x^2}

where:

<em>Ep = elastic potential energy ( J )</em>

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Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

mass of object = m = 1.25 kg

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final extension = x' = 0.0735 - 0.0275 = 0.0460 m

<u>Asked:</u>

kinetic energy = Ek = ?

<u>Solution:</u>

<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>

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\texttt{ }

<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>

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Ek = \frac{1}{2}k ( (x')^2 - x^2 ) + mgh

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\boxed {Ek \approx 1.20 \texttt{ J}}

\texttt{ }

<h3>Learn more</h3>
  • Kinetic Energy : brainly.com/question/692781
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  • Young Modulus : brainly.com/question/9202964
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\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

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