Tangential and Normal Components of Acceleration | Query Point Official

Tangential and Normal Components of Acceleration

Problem Statement

The particle moves along the curve:

y^2 – 2x = 1 – 2y with uniform speed v. Find the tangential and normal components of acceleration at the point:

(0, -1 ± √2)

Solution Overview

We calculate:

• Tangential acceleration: `a_T = dv/dt`
• Normal acceleration: `a_N = v^2 / ρ`

Step 1: Tangential Component

Since speed is uniform:

`a_T = dv/dt = 0`

Step 2: Normal Component

Normal acceleration formula:

`a_N = v^2 / ρ`, where `ρ = [(1 + (dy/dx)^2)/|d^2y/dx^2|]^(3/2)`

Step 3: Derivatives

First derivative of the curve:

`2y (dy/dx) – 2 = -2 (dy/dx)`

Simplified:

`dy/dx = 1 / (y + 1)`

At point (0, -1 ± √2): `dy/dx = 1 / (±√2)`

Second derivative:

`d^2y/dx^2 = -1 / (y + 1)^3`

At the point (0, -1 ± √2): `d^2y/dx^2 = -1 / (±√2)^3`

Step 4: Radius of Curvature

Using formula:

`ρ = [(1 + (1/(±√2))^2) / (-1 / (±√2)^3)]^(3/2)`

Simplified:

`ρ = [3 (±√2)]^(3/2)`

Step 5: Normal Component

Finally, the normal component:

`a_N = v^2 / [3 (±√2)]^(3/2)`

Conclusion

The tangential acceleration is zero due to constant speed. The normal acceleration depends on the curvature of the curve at the given point. This problem demonstrates the application of calculus and dynamics in computing acceleration components for a particle moving along a curve.

FAQ

Q1: Why is tangential acceleration zero?

Because the particle moves at uniform speed, so `dv/dt = 0`.

Q2: How to compute radius of curvature?

Use `ρ = [(1 + (dy/dx)^2)/|d^2y/dx^2|]^(3/2)` and plug in the derivatives at the point to find the curvature radius.

Q3: Are there similar problems?

See Mathematics Notes & MCQs for similar problems in Calculus and Dynamics. This strengthens understanding of tangential and normal acceleration concepts.