Rolling Motion Explained: Cycloid Path & Displacement of a Wheel Point

 - Dr.Sanjaykumar Pawar   

Physics Notes: Displacement in Rolling Motion of a Wheel (Cycloid Concept)

Cycloid path traced by a point on a rolling wheel during half a revolution.

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NEET Physics: Rolling Motion (Questions with Step-by-Step Solutions)

Q1. A wheel of radius 2 m rolls through angle π. Find displacement of center.

Step 1: x = rθ
Step 2: x = 2 × π = 2π
Answer: 2π m

Q2. Vertical displacement of a point after half rotation.

Step 1: Bottom → Top of wheel
Step 2: Height = 2r
Answer: 2r

Q3. Radius = 1 m, θ = π. Find displacement of point initially at contact.

Step 1: x = r(θ - sinθ), y = r(1 - cosθ)
Step 2: sinπ = 0, cosπ = -1
x = π, y = 2
Step 3: D = √(x² + y²)
D = √(π² + 4)
Answer: √(π² + 4)

Q4. Path traced by a point on rim of rolling wheel.

Answer: Cycloid

Q5. Radius 3 m, θ = 2π. Find displacement of center.

Step 1: x = rθ
Step 2: x = 3 × 2π = 6π
Answer: 6π m

Q6. Radius 3 m, half rotation. Find displacement of point initially at contact.

Step 1: x = 3π, y = 6
Step 2: D = √(9π² + 36)
D = 3√(π² + 4)
Answer: 3√(π² + 4)

Q7. Radius 2 m, half rotation. Find center displacement.

x = rθ = 2 × π = 2π
Answer: 2π m

Q8. Maximum height of point on rolling wheel.

Height = 2r
Answer: 2r

Q9. Velocity of contact point in pure rolling.

Instantaneous velocity = 0
Answer: Zero

Q10. Path of center of rolling wheel.

Answer: Straight line

Question

A wheel of radius 3 m rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is:

1. 2π m
2. √(2π) m
3. √(π² + 4) m
4. 3√(π² + 4) m

Solution (Step by Step)

Step 1: Given data

• Radius of wheel, r = 3 m
• Rotation = half revolution = π radians

Step 2: Motion of point on rolling wheel

A point on the rim of a rolling wheel follows a cycloid path.

Position equations:

x = r(θ − sinθ),    y = r(1 − cosθ)

Step 3: Substitute values (θ = π)

• sin π = 0
• cos π = -1

x = 3(π − 0) = 3π
y = 3(1 − (−1)) = 6

Step 4: Displacement

Displacement = √(x² + y²)

= √((3π)² + 6²)
= √(9π² + 36)
= 3√(π² + 4)

Final Answer

Option (4): 3√(π² + 4) m

Short Notes: Rolling Motion

1. Rolling without slipping

Condition: v = rω. The point of contact is instantaneously at rest.

2. Cycloid motion

A point on the rim of a rolling wheel traces a cycloid curve.

3. Key equations

x = r(θ − sinθ),    y = r(1 − cosθ)

4. Half revolution result

For θ = π: horizontal shift = rπ, vertical rise = 2r.

5. Important concept

Displacement depends on both horizontal and vertical motion, not just rolling distance.

Short Notes: Rolling Motion and Displacement of a Point on a Wheel

1. Rolling without slipping

When a wheel rolls without slipping, the point of contact with the ground is momentarily at rest. The linear speed of the center of the wheel is related to angular speed by:

v = rω

2. Motion of a point on the rim (Cycloid)

A point on the rim of a rolling wheel traces a curve called a cycloid. The position of a point initially at the ground contact after rotation θ is:

x = r(θ − sinθ)
y = r(1 − cosθ)

3. Displacement in rolling motion

Displacement is the straight-line distance between initial and final positions of the point. It is found using:

Displacement = √(x² + y²)

4. Half revolution case

For θ = π:

• Horizontal shift = rπ
• Vertical height = 2r

The point rises to the top of the wheel.

5. Key idea from the problem

Even though the wheel moves horizontally, a point on its rim follows a curved path. The final displacement depends on both horizontal and vertical motion, not just horizontal rolling distance.

6. Important takeaway

Rolling motion combines rotation and translation. Cycloid equations are the standard tool for analyzing such motion.