Question

A 30 kg child is riding on a Ferris Wheel that has a radius of 15.0 m. The Ferris Wheel turns in a vertical circle once every 10.0 sec-onds. Find the magnitude of the normal force the seat applies to the child at the top and at

the bottom ot the ride.

Answer 1

The problem involves calculating the normal force exerted on a child by a Ferris wheel seat at both the highest and lowest points of the ride, using principles of circular motion and centripetal force.

Step-by-step explanation:

The question involves calculating the normal force exerted by a Ferris wheel seat on a child at the top and bottom of the ride using concepts of centripetal force and circular motion. To find these forces, we first need to calculate the centripetal acceleration of the child as the Ferris wheel rotates.

Centripetal acceleration is given by the formula a = v^2 / r, where v is the tangential velocity and r is the radius of the circle. Since the Ferris wheel completes one rotation every 10 seconds, the tangential velocity v can be found using the circumference of the circle (C = 2πr) and the time period (T = 10 sec). Hence, v = C/T = 2π(15.0 m) / 10.0 s.

Once we have the centripetal acceleration, we can calculate the normal force at the top and bottom of the Ferris wheel using Newton's second law. At the top, the normal force and the gravitational force both point towards the center of the Ferris wheel, thus N + mg = ma, where m is the mass of the child and g is the acceleration due to gravity. At the bottom, the normal force points away from the center, so N - mg = ma. We can solve for N in both cases to find the normal forces at the top and bottom.

Going through the calculations, we'll determine the magnitude of the normal force required to keep the child in motion on the Ferris wheel both at the highest and lowest points of the ride.

Answer 2

Step-by-step explanation:

We can use the following equation to calculate the normal force at the top and bottom of the ride:

N = mg ± mv^2/r

where N is the normal force, m is the mass of the child (30 kg), g is the acceleration due to gravity (9.81 m/s^2), v is the velocity of the child, r is the radius of the Ferris wheel (15.0 m), and the ± sign indicates that we need to use the plus sign for the bottom of the ride and the minus sign for the top of the ride.

At the bottom of the ride, the child is moving in the same direction as the force of gravity, so we use the plus sign:

Nbottom = mg + mv^2/r

The velocity of the child can be calculated using the formula for centripetal acceleration:

a = v^2/r

We can rearrange this formula to solve for v:

v = sqrt(ar)

where a is the centripetal acceleration (which is equal to g at the bottom of the ride), so:

v = sqrt(gr) = sqrt(9.81 m/s^2 × 15.0 m) = 11.2 m/s

Now we can substitute the values into the equation for Nbottom:

Nbottom = mg + mv^2/r = (30 kg)(9.81 m/s^2) + (30 kg)(11.2 m/s)^2/15.0 m = 536 N

Therefore, the normal force at the bottom of the ride is 536 N.

At the top of the ride, the child is moving in the opposite direction as the force of gravity, so we use the minus sign:

Ntop = mg - mv^2/r

The velocity of the child at the top of the ride is the same as at the bottom, so we can use the same value of v:

Ntop = mg - mv^2/r = (30 kg)(9.81 m/s^2) - (30 kg)(11.2 m/s)^2/15.0 m = 344 N

Therefore, the normal force at the top of the ride is 344 N.