Calculating Centripetal Force on a Child Riding a Ferris Wheel

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A 40kg child on a Ferris wheel with a radius of 9m and a rotation speed of 4 times per minute experiences specific forces at different points. The angular velocity is calculated at 0.419 radians per second, with a tangential velocity of 3.77 m/sec and centripetal acceleration of 1.58 m/sec². At the lowest point, the normal force exerted by the seat combines gravitational and centripetal forces, while at the highest point, it subtracts the centripetal force from gravity. The equations for calculating the normal force at both points are N = mg + mv²/r for the bottom and N = mg - mv²/r for the top. The calculations confirm the correct application of centripetal force principles in this scenario.
leisiminger
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A 40kg child takes a ride on a Ferris Wheel that rotates 4 times each minute and has a radius of 9m.

I firgured that angular velocity is .419 rads/sec
I figured that tangential velocity is 3.77m/sec
I figured that centripetal acceleration is 1.58m/sec^2

Questions
a)What force in N does the seat exert on the child at the lowest point of the Ferris wheel?

b)What force in N does the seat exert on the child at the highest point of the Ferrish Wheel? Keep in mind that radially inward in the positive direction
 
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leisiminger said:
A 40kg child takes a ride on a Ferris Wheel that rotates 4 times each minute and has a radius of 9m.

I firgured that angular velocity is .419 rads/sec
I figured that tangential velocity is 3.77m/sec
I figured that centripetal acceleration is 1.58m/sec^2

Questions
a)What force in N does the seat exert on the child at the lowest point of the Ferris wheel?

b)What force in N does the seat exert on the child at the highest point of the Ferrish Wheel? Keep in mind that radially inward in the positive direction

Draw a force diagram at the bottom and the top.

F = m*a - so whatever acceleration the mass is experiencing should translate to the seat shouldn't it?
 
i figure for the bottom its
N= mg +mv^2
-------------
r

and the top would be

N= mg - mv^2
----------------
r
 
leisiminger said:
i figure for the bottom its
N= mg +mv^2
-------------
r

and the top would be

N= mg - mv^2
----------------
r

I think that's right, but your post displays unclearly.

N = m*g + m*v2/r

So if that is your first one and similarly for the second ... yes.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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