Linear and Angular Variables

4.1.5 Relationship Between Linear and Angular Quantities

29. Linear and Angular Displacement

The relation between arc length and angular displacement is:

s = rθ

Therefore:

• Larger radius produces greater linear displacement.
• Angular displacement remains the same for every point of a rigid body.

30. Linear and Angular Velocity

Starting from:

s = rθ

Dividing by time:

st = r (θt)

Therefore:

v = rω

The velocity v is tangent to the circular path.

Points farther from the rotational axis have greater tangential speeds for the same angular velocity.

31. Tangential Acceleration

From:

v = rω

The rate of change of linear speed is related to the rate of change of angular velocity:

a_t = rα

Tangential acceleration changes the magnitude of linear velocity.

It acts tangent to the circular path.

32. Radial Acceleration

Radial acceleration is directed toward the centre of the circular path.

It arises because the direction of velocity changes.

It is also called centripetal acceleration:

a_c = v²r

Using v = rω:

a_c = (rω)² / r a_c = rω²

Centripetal acceleration is explained fully in the next textbook topic.

33. Total Linear Acceleration

Tangential and radial accelerations are perpendicular.

In vector form:

a⃗ = a⃗_t + a⃗_c

Its magnitude is:

a = √(a_t² + a_c²)

Tangential acceleration changes speed.

Radial acceleration changes direction.

34. Important Rigid-Body Rule

For all particles in the same rigid rotating body:

• Angular displacement θ is the same.
• Angular velocity ω is the same.
• Angular acceleration α is the same.

However, linear quantities depend on radius:

s = rθ v = rω a_t = rα

Therefore, points farther from the axis move faster and travel farther.

35. Book-Based Solved Example: Rolling Tyre

Question: A bicycle tyre of radius 0.331 m rolls without slipping with an initial linear speed of 6.80 m s⁻¹. It stops after travelling 74.8 m. Find: Initial angular velocity Number of revolutions Angular acceleration Stopping time

Given

r = 0.331 m v_i = 6.80 m s⁻¹ ω_f = 0 s = 74.8 m

Initial Angular Velocity

v_i = r ω_i ω_i = v_fracir ω_i = 6.800.331 ω_i = 20.54 rad s⁻¹

Number of Revolutions

Distance covered in one revolution is:

2πr

Therefore:

N = s2πr N = 74.8 / (2π(0.331)) N = 35.96 rev

Angular Displacement

Δθ = 2πN Δθ = 2π(35.96) Δθ ≈ 225.94 rad

Angular Acceleration

ω_f² = ω_i² + 2α Δθ (0)² = (20.54)² + 2α(225.94) 0 = 421.89 + 451.88α α = −421.89451.88 α = −0.934 rad s⁻²

Stopping Time

ω_f = ω_i + αt 0 = 20.54 + (−0.934)t 0 = 20.54 − 0.934t 0.934t = 20.54 t = 21.99 s

Final Answers

ω_i = 20.54 rad s⁻¹ N ≈ 36.0 rev α = −0.934 rad s⁻² t ≈ 22.0 s