Vertical Circular Motion of a giant wheel

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SUMMARY

The discussion centers on the vertical circular motion of a giant wheel with a diameter of 40 meters, focusing on the forces acting on a man standing on a platform at the top of the wheel. The key equation derived is T + mg = mv²/r, where T represents the normal force, m is the mass of the man, g is the acceleration due to gravity (9.81 m/s²), and r is the radius of the wheel (20 meters). The solution involves calculating the normal force and applying Newton's laws to determine the speed of the man at the top of the wheel.

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c.melissas
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1. A giant wheel, 40 meters in diameter, is fitted with a cage and platform on which a man can stand. The wheel rotates at such a speed that when the cage is at the top of the wheel, then the force exerted by the man on the platform is equal to his weight. The speed of the man is:

2. T + mg = mv^2 / r and in this specific problem, v = m.

3. T + m(9.81) = mv^2 / 40
40(T + m(9.81)) = mv^2
Square root of [ (40(T + m(9.81))) / m ]

I think I am on the right track, but I am not really sure what do from this point.
 
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Can you draw a free body diagram?

40m is the diameter.
 
First there is not a tension force really, it is more of a normal force since it is applied from the platform. That being said the key idea involves using Newtons law that forces are equal and opposite. If the platform feels a force of W by the man then what force is applied on the man by the platform? This is your normal force N. Then sum the forces in the normal direction to get N+W=mv^2/r. As mentioned above determine your normal force and the problem is solved.
 

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