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Nadusha1986 [10]
3 years ago
15

A child of mass 47 kg sits on the edge of a merry-go-round with radius 1.3 m and moment of inertia 56.3953 kg m2 . The merrygo-r

ound rotates with an angular velocity of 1.6 rad/s. What radial force does the child have to exert to stay on the merry-go-round?
Physics
1 answer:
lakkis [162]3 years ago
4 0

Answer:

F=156.416\ N the child have to exert this much amount of force radially to stay on the wheel.

Explanation:

Given:

mass of the child, m=47\ kg

radius of the merry-go round wheel, r=1.3\ m

moment of inertia of the wheel, I=56.3953\ kg.m^2

angular velocity of the wheel, \omega=1.6\ rad.s^{-1}

Since the child is sitting on the edge of the rotating wheel the child will feel and outward throwing force called centrifugal force.

Centrifugal force is mathematically given as:

F=m.r.\omega^2

F=47\times 1.3\times 1.6^2

F=156.416\ N the child have to exert this much amount of force radially to stay on the wheel.

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The new magnitude of the force of attraction will be 6 times the original force of attraction

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<h3>How to determine the new force </h3>
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But

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What are the rules for setting up an integral of rotation?
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A Jeep accelerated at 2.2 m/s2 until after 18 seconds its displacement was 660 meters. Assuming the Jeep traveled in a straight
pogonyaev
Ok so we know:
The time (t) is 18seconds
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