A Ferris wheel with a radius of 30 m spins rapidly as part of a malfunction. If the Ferris wheel completes five rotations in one minute and each passenger car has a mass of 300 kg, what is the centripetal force of one passenger car at the top of the Ferris wheel?

The centripetal force of one passenger car with a mass of 300 kg at the top of the Ferris wheel with a radius of 30 m is 468 N.

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A Ferris wheel problem is directly related to the centripetal acceleration. Depending upon the position of the passenger car, the riders may feel lighter or heavier. In the given problem, we are asked to find the centripetal force of the passenger car at the top of the Ferris wheel.

Please refer to the attached figure. C is the position of the passenger car, R is the radius of the Ferris wheel, and `\omega` is the angular velocity of the Ferris wheel in rad/s.

The forces acting on the passenger car at the top of the Ferris wheel are due to the combined effect of gravity and centripetal acceleration caused by the rotation of the Ferris wheel with angular velocity `\omega` .

Now let `F_c` be the force exerted by the passenger car at point C (i.e., at the top of the Ferris wheel). `F_g ` is the force of gravity acting downward where `F_g = mg `

Given that the mass of the car is 300 kg and we know that `g = 9.8 m/s^2`

So, we get `F_g = 300 \times 9.8 = 2940 \ kg\ m/s^2 = 2940 N`

Now let `a` be the centripetal acceleration at point C. Acceleration always points toward the center of the Ferris wheel. So `a` is pointing down at point C. Hence it would be negative. Mathematically, centripetal acceleration is given by:

`a = \omega ^2 R`

It is given that the Ferris wheel completes 5 rotations in one minute (60 seconds). One complete rotation is `2\pi ` radians . So 5 complete rotations means `5\times 2\pi =10\pi` radians. Hence angular velocity is `\omega = \frac{10\ pi}{60} = 0.524 ` rad/s

The radius of the Ferris wheel is given as R=30 m

Therefore,

`a = (0.524)^2\times 30 = 8.24\ m/s^2 `

Now by Newton's second law,

`\sum F = ma`

where F is the sum of all forces acting on the passenger car at point P.

`\sum F = F_c -F_g`

Therefore we can write,

`F_c-F_g = m(-a)` (a is negative since it is pointing down toward the center of the wheel)

So,

`F_c = F_g -ma `

`= 2940 - (300 \times 8.24)`

`= 2940 - 2472`

`=468\ N`

Hence the centripetal force of one passenger car with a mass of 300 kg at the top of the Ferris wheel with a radius of 30 m is 468 N.

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