Suppose you and a friend, each of mass 65 kg, go to the park and get on a 4.5 m diameter merry-go-round, while your friend pushes so that it rotatesonce every 6.5 seconds. What is the magnitude of...

Suppose you and a friend, each of mass 65 kg, go to the park and get on a 4.5 m diameter merry-go-round, while your friend pushes so that it rotates

once every 6.5 seconds. What is the magnitude of the (apparent) outward force you feel?

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jeew-m's profile pic

jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

Rotational speed = 1rev/6.5 S

1 rev = 2*pi rad

Therefore rotational speed (w) = 2*pi/(6.5) = 0.967 rad/s

Radius of the circle = Diameter/2 = 4.5/2 = 2.25

Since you are rotating around the center of the merry-go-round a centrifugal force will act towards the center of the merry-go-round.

This centrifugal force = m*r*w^2

Where       m= mass of the rotating item in kg

                 r= radius of the circle in m

               w = angular velocity in rad/s

 

According to the Newton’s law every action has an equal and opposite reaction. Due to centrifugal force you will feel the opposite of that as an outward force which has the same magnitude as centrifugal force.

Centrifugal force              = m*r*w^2

                                     = 65*(4.5/2)*(0.967) ^2

                                     = 136.756 N

 

So the outward force = 136.756 N

Sources:
najm1947's profile pic

najm1947 | Elementary School Teacher | (Level 1) Valedictorian

Posted on

The rotational speed = 1 revolution per 6.5 second = 2pi/6.5 radians/second

Diameter of the merry-go-round = 4.5m

Radius = 4.5/2 =2.25m

Centrifugal force on each = mrw^2, where m = mass, r = redius and w = rotational speed

Force = 65*2.25*(2pi/6.5)^2 = 136.7N

The magnitude of (apparent) outward force is 136.7N

 

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