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masha68 [24]
3 years ago
8

This problem describes an experimental method for determining the moment of inertia of an irregular shaped object such as the pa

yload for a satellite. A counter weight of mass m, is suspended by a cord wound around a spool of radius r, forming part of a turntable supporting the object. The turntable can rotate w/o friction. When the counterweight is released from rest it descends through a distance h, acquiring a speed v. Show that the moment of inertia I of the rotating apparatus is mr2 (2gh/v2 – 1).
Physics
1 answer:
Scorpion4ik [409]3 years ago
8 0

Answer:

Explanation:

 The turntable that is rotating apparatus whose moment of inertia ( I )  is to be calculated is used as a spool of radius r. m mass is going down with acceleration a and rotating the apparatus with the help of cord with tension T  wound around it

For motion of weight m

mg - T = ma

and for the motion of spool or apparatus

Tr = I a/r

T = Ia/r²

mg - Ia/r²  = ma

mg   =  Ia/r² + ma

a (  I/r² + m ) = mg

When  the counterweight m  is released from rest it descends through a distance h, acquiring a speed v

v² = 2ah

a = v² / 2h

v² / 2h  (  I/r² + m ) = mg

(  I/r² + m ) = 2mgh / v²

 I/r²  = 2mgh / v²-  m

I = r² (2mgh / v²-  m )

= mr² ( 2gh/ v² -1 )

Proved .

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Answer:

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Explanation:

Given:

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Radius of the merry-go-round, r = 1.17~m

the force on the merry-go-round, F = 57.8~N

Acceleration due to gravity, g= 9.8~m.s^{-2}

Time given, t=3.47~s

Mass of the merry-go-round is given by

m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg

Moment of inertial of the merry-go-round is given by

I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}

Torque on the merry-go-round is given by

\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m

The angular acceleration is given by

\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}

The angular velocity is given by

\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}

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Answer:

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Explanation:

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\displaystyle \epsilon(t) = \frac{d}{dt}(\Phi(t)).

Finding the average EMF in the coil is similar to finding the average velocity.

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However, by the Fundamental Theorem of Calculus, integration reverts the action of differentiation. That is:

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