A 40 kg child rides on the edge of a merry-go-turning with a rotational speed of

Question

A 40 kg child rides on the edge of a merry-go-turning with a rotational speed of 25 revolutions/min (rpm). The merry-go-round has a diameter of 6 meters. Determine the rotational speed of the merry-go-round in SI units of radians/sec. Determine the acceleration of the child. Express your final answer in units of g's. Would you consider this magnitude of acceleration to be safe for a child? If the rotational of the merry-go-round were doubled, would this cause the acceleration to be increased by a factor of 27 Briefly explain/support your answer?

Explanation / Answer

The force will be created by the vector addition of that needed to counter gravitational and provide centripetal accelerations acting on the child's mass.

The vertical acceleration will equal gravity g = 9.81 m/s² as tangential velocity is constant

the centripetal acceleration will be horizontal to the right at given position and is found with the formula
ac = ²r

= 25 rev/min (2 rad/rev) / 60 sec/min = 5/6 rad/s = 2.61799 rad/s ..........................ans a)

ac = (5/6 )²x6.0
acceleration of the child

ac = 41.12 m/s² = 4.19 g ...................................ans b)

no this acceleration is not safe for the hcild.. it s a huge value.

now if the rotational speed of the merry-go-round is doubled then

omega = 2x2.61799 = 5.23598 rad/s

so, acceleration = (5.23598)^2 x6.0 = 164.4929 m/s^2 = 4x first acceleration

no the acceleration become four times only

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