A child stand on the rim of merry go round of

a child stand on the rim of merry go round of radius 2m rotating at 10 rpm. determine the minimum coefficient friction between her feet and the floor to keep her in circular motion

Answer

Minimum Coefficient of Friction for Circular Motion on a Merry-Go-Round

When a child stands at the edge of a rotating merry-go-round, their body experiences a tendency to move in a straight line due to inertia. To remain on the circular path, a force must act towards the center of the circle to keep them rotating with the platform. This force is known as the centripetal force, and it is provided by the friction between their feet and the floor of the rotating surface.

🌪️ Physics Behind the Scenario

The concept of circular motion involves several important quantities:

Centripetal Acceleration (a_c) – the acceleration directed towards the center of the circle.
Centripetal Force (F_c) – the net force required to keep the body moving in a circle.
Frictional Force – the static friction that supplies the required centripetal force in this case.

Centripetal Force: F_c = m × v² / r

In this problem, the child is not sliding or accelerating vertically, so the frictional force must equal the required centripetal force. The maximum static friction is:

F_friction = μ × N = μ × m × g

To prevent slipping:

μ × m × g ≥ m × v² / r

Cancelling mass (m) from both sides:

μ ≥ v² / (r × g)

🧾 Given Data

Radius of merry-go-round, r = 2 m
Rotational speed = 10 rpm (revolutions per minute)
Acceleration due to gravity, g = 9.8 m/s²

🔁 Step 1: Convert Rotational Speed to Angular Velocity

First, convert rpm to angular velocity in radians per second:

ω = 2π × (rpm) / 60 = 2π × 10 / 60 = π / 3 ≈ 1.047 rad/s

➡️ Step 2: Find Linear Velocity

Use the relation between linear velocity and angular velocity:

v = ω × r = 1.047 × 2 = 2.094 m/s

✳️ Step 3: Plug into the Friction Formula

Now, calculate the minimum coefficient of friction:

μ ≥ (2.094)² / (2 × 9.8) ≈ 4.387 / 19.6 ≈ 0.224

✅ Final Result

The minimum coefficient of static friction required between the child’s feet and the merry-go-round floor to keep her in circular motion is:

μ_min ≈ 0.224

📊 Practical Meaning

This means that any frictional coefficient less than 0.224 would be insufficient to hold the child in circular motion, and she would slip outward due to inertia. Surfaces like dry rubber-soled shoes on a metal or wooden platform often exceed this value, ensuring safety. However, wet or dusty conditions could reduce the coefficient and lead to slipping.

📚 Important Concepts Recap

Circular motion requires a constant inward (centripetal) force.
Friction is the only force providing this inward pull for a standing person.
Minimum coefficient of friction can be calculated using linear velocity and radius.

🧠 Related Real-World Examples

Children on spinning rides at amusement parks
Passengers turning in vehicles
Objects placed on rotating turntables or carousels
Earth’s rotation causing pseudo-forces at the equator

🔎 Test Your Understanding

Try calculating the required coefficient of friction if:

The radius is increased to 3 m
The merry-go-round spins at 15 rpm

Use the same method: convert rpm to angular velocity, find linear velocity, and apply it in: