Math Problem Statement

A child with a mass of 44kg sits on a Ferris wheel with radius 8m rotating with constant speed v.

At the highest point, the magnitude of the normal force n on the student exerted by the seat on which she is sitting is 341N.

What is the speed (m/s) with which the Ferris wheel turns? g is 10

Solution

To solve this, we can apply Newton’s second law in the vertical direction at the highest point of the Ferris wheel. The forces acting on the child at the highest point are gravity and the normal force. We can set up the following equation:

Given Data

  • Mass of child, m=44kgm = 44 \, \text{kg}
  • Radius of Ferris wheel, r=8mr = 8 \, \text{m}
  • Gravitational acceleration, g=10m/s2g = 10 \, \text{m/s}^2
  • Normal force at the highest point, n=341Nn = 341 \, \text{N}

Step 1: Write the Force Equation

At the highest point, the centripetal force is provided by the difference between the gravitational force and the normal force. Thus,

mgn=mv2rmg - n = \frac{mv^2}{r}

where:

  • mgmg is the gravitational force acting on the child.
  • nn is the normal force exerted by the seat on the child.
  • mv2r\frac{mv^2}{r} is the centripetal force required to keep the child moving in a circular path.

Step 2: Substitute Values

Calculate mgmg:

mg=44×10=440Nmg = 44 \times 10 = 440 \, \text{N}

Now substitute mg=440Nmg = 440 \, \text{N}, n=341Nn = 341 \, \text{N}, m=44kgm = 44 \, \text{kg}, and r=8mr = 8 \, \text{m} into the equation:

440341=44×v28440 - 341 = \frac{44 \times v^2}{8}

This simplifies to:

99=44×v2899 = \frac{44 \times v^2}{8}

Multiply both sides by 8:

792=44v2792 = 44v^2

Now, divide by 44:

v2=79244=18v^2 = \frac{792}{44} = 18

Take the square root of both sides:

v=184.24m/sv = \sqrt{18} \approx 4.24 \, \text{m/s}

Final Answer

The speed vv with which the Ferris wheel turns is approximately 4.24 m/s.


Would you like further details, or do you have any questions?

Here are five related questions for practice:

  1. What would be the normal force at the lowest point of the Ferris wheel?
  2. How would the speed change if the mass of the child were increased?
  3. What would happen to the normal force if the speed increased?
  4. How does the centripetal force relate to the gravitational force in circular motion?
  5. How would the problem change if the Ferris wheel radius increased?

Tip: In circular motion, centripetal force always points toward the center and depends on the square of the speed divided by the radius.

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Math Problem Analysis

Mathematical Concepts

Circular Motion
Centripetal Force
Newton's Second Law

Formulas

mg - n = (mv^2)/r
v = sqrt((mg - n) * r / m)

Theorems

Newton's Second Law of Motion

Suitable Grade Level

Grades 9-12