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sveta [45]
3 years ago
15

A horizontal 826 N merry-go-round of radius 1.17 m is started from rest by a constant horizontal force of 57.8 N applied tangent

ially to the merry-go-round. Find the kinetic energy of the merry-goround after 3.47 s. The acceleration of gravity is 9.8 m/s 2 . Assume the merry-go-round is a solid cylinder. Answer in units of J. 028 10.0
Physics
1 answer:
Julli [10]3 years ago
5 0

Answer:

The kinetic energy of the merry-go-round is \bf{475.47~J}.

Explanation:

Given:

Weight of the merry-go-round, W_{g} = 826~N

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}

The kinetic energy of the merry-go-round is given by

E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J

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