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2.3 · Use normal and tangential components along a curved path

Learn to use normal and tangential components along a curved path through clear examples and targeted practice.

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Curvilinear and Relative Motion

Separate acceleration caused by changes in speed from acceleration caused by changes in direction

A particle can accelerate while moving at constant speed if its direction changes. If its speed changes too, acceleration has another component. Normal and tangential components let us study these effects separately. In this lesson, the system is one particle, observed from a stationary frame fixed to the ground. The path lies in a plane. At the instant of interest, positive tangential direction is along the particle’s motion, and positive normal direction points inward toward the local centre of curvature. These local directions move with the particle as it travels.

What you will learn

  • Define tangent and inward normal directions at a particle’s position on a curved path.
  • Use normal–tangential acceleration equations with consistent signs and SI units.
  • Apply Newton’s second law in normal and tangential directions to find motion or force quantities.

1. Define the local directions

At a point on a smooth curved path, the tangent direction follows the instantaneous velocity. Let its unit vector be et\mathbf{e}_t. The inward normal is perpendicular to the tangent and points toward the local centre of curvature; let its unit vector be en\mathbf{e}_n. The distance from the particle to that centre is the radius of curvature, ρ\rho.
Position specifies where a particle is, while path length is the distance it has travelled along its route. Displacement is the vector change in position. Velocity is the rate of change of position and is tangent to the path. Acceleration is the rate of change of velocity. It can result from a change in speed, a change in direction, or both.
The tangential acceleration is the rate at which speed changes. With positive tangent chosen along the motion, it is positive when the particle speeds up and negative when it slows down. The normal acceleration points inward and has magnitude v2/ρv^2/\rho, where vv is speed. Thus the acceleration vector is the sum of perpendicular tangential and normal components.
a=v˙ et+v2ρ en\mathbf{a}=\dot v\,\mathbf{e}_t+\frac{v^2}{\rho}\,\mathbf{e}_n
  • The tangent follows velocity; the inward normal points toward the local centre of curvature.
  • Tangential acceleration changes speed; normal acceleration changes velocity direction.
  • For a straight path, curvature is zero and normal acceleration is zero.

2. Apply Newton’s second law along the path

Forces can be resolved along the tangent and inward normal, just as they can be resolved along horizontal and vertical axes. Applying Newton’s second law separately in these perpendicular directions gives the tangential and normal equations. The sum includes all actual forces acting on the particle, resolved with the chosen signs.
Here, a positive tangential force component points along the motion, and a positive normal force component points inward. A force directed outward therefore contributes negatively to the normal sum. If the solved tangential acceleration is negative, the particle is slowing down under our direction convention.
Do not confuse normal acceleration with a surface’s normal force. Normal acceleration describes motion perpendicular to the path. A surface normal force is an actual force that may contribute to the inward resultant, along with forces such as gravity, tension, or friction. Identify the system and frame, sketch the path and forces, mark the local directions, list knowns and unknowns, and then resolve forces consistently.
∑Ft=mv˙,∑Fn=mv2ρ\sum F_t=m\dot v,\qquad \sum F_n=m\frac{v^2}{\rho}
  • Use Newton’s second law separately along tangent and inward normal.
  • The normal direction is perpendicular to the path, not necessarily vertical.
  • A negative tangential result means the particle is slowing for the chosen direction of travel.

3. Check dimensions, directions, and limiting cases

The normal acceleration has units (m/s)2/m=m/s2(\mathrm{m/s})^2/\mathrm{m}=\mathrm{m/s^2}, the same as tangential acceleration. Multiplying either component by mass gives force units, since kg m/s2=N\mathrm{kg\,m/s^2}=\mathrm{N}.
At constant speed, v˙=0\dot v=0, but a moving particle on a curve still has inward normal acceleration. On a straight segment, the radius of curvature becomes unbounded, so the normal acceleration tends to zero for finite speed. At rest, the expression for normal acceleration gives zero. For a circular path, the radius of curvature is the circle’s radius.
The magnitude of total acceleration combines the perpendicular components. Do not add their magnitudes directly. Check that each component’s sign matches its chosen direction and that the final result has the correct dimensions.
a=at2+an2a=\sqrt{a_t^2+a_n^2}
  • Constant speed removes tangential acceleration, not normal acceleration.
  • A straight path has zero normal acceleration.
  • Combine perpendicular acceleration components using the Pythagorean relation.

Worked example

Constant-speed motion on a circular path

A 900 kg car travels at a constant speed of 12 m/s around a level circular track of radius 60 m. Find its acceleration and the inward horizontal resultant force required. The car’s path is explicitly circular.
Car on a circular path
Car on a circular pathxy60 m radiuscar12 m/s

At the right side of the circular path, the car’s counterclockwise velocity is upward and tangent to the path.

  1. Define the system and directions
    The system is the car, and the observer is fixed to the ground. At the instant considered, positive tangent follows the car’s motion and positive normal points toward the circle’s centre. The initial state is motion at 12 m/s on the circle; the acceleration and resultant force at this instant are sought.
  2. Identify knowns and model
    The speed is constant, so its rate of change along the path is zero. For a circular path, the radius of curvature is the circle’s radius. Use normal–tangential acceleration and Newton’s second law; the inward resultant force is not necessarily one individual force.
    v˙=0,ρ=60 m\dot v=0,\qquad \rho=60\ \mathrm{m}
  3. Calculate acceleration
    The tangential component vanishes. The normal component points inward and has magnitude equal to speed squared divided by radius.
    at=0,an=(12 m/s)260 m=2.40 m/s2a_t=0,\qquad a_n=\frac{(12\ \mathrm{m/s})^2}{60\ \mathrm{m}}=2.40\ \mathrm{m/s^2}
  4. Find the required resultant force
    Apply Newton’s second law in the inward normal direction. Multiplying the mass by the inward acceleration gives the required inward resultant force.
    ∑Fn=(900 kg)(2.40 m/s2)=2.16×103 N\sum F_n=(900\ \mathrm{kg})(2.40\ \mathrm{m/s^2})=2.16\times10^3\ \mathrm{N}
Answer: The acceleration is 2.40 m/s² inward, with zero tangential component. The required inward horizontal resultant force is 2.16 kN.
Check: The force units are kg·m/s², or N. The direction is inward, as required for circular motion. Constant speed is consistent with zero tangential acceleration.

Worked example

Speeding up while following a circular arc

A particle follows a circular arc of radius 8.0 m. At one instant its speed is 6.0 m/s and is increasing at 1.5 m/s². Find its acceleration components and magnitude. The path is counterclockwise, and the particle’s mass is 2.0 kg.
Speed increasing on an arc
Speed increasing on an arcxy8.0 m radiusparticle6.0 m/s

At the right side of the circular path, velocity and positive tangential acceleration point upward; inward normal acceleration points toward the centre.

  1. Set the frame and local axes
    The system is the particle, observed from a stationary ground frame. Positive tangent points along its counterclockwise motion; positive normal points toward the circle’s centre. The stated instant is the state for which acceleration is sought.
  2. Determine the components
    The given rate of speed increase is positive tangential acceleration. The normal component follows from the instantaneous speed and radius of curvature.
    at=1.5 m/s2,an=(6.0 m/s)28.0 m=4.5 m/s2a_t=1.5\ \mathrm{m/s^2},\qquad a_n=\frac{(6.0\ \mathrm{m/s})^2}{8.0\ \mathrm{m}}=4.5\ \mathrm{m/s^2}
  3. Combine perpendicular components
    The tangent and inward normal are perpendicular, so use the magnitude relation. The positive tangential value means the particle is speeding up; the normal component points inward.
    a=(1.5 m/s2)2+(4.5 m/s2)2=4.74 m/s2a=\sqrt{(1.5\ \mathrm{m/s^2})^2+(4.5\ \mathrm{m/s^2})^2}=4.74\ \mathrm{m/s^2}
  4. Check with Newton’s second law
    As a force check, multiply each acceleration component by the particle’s mass. The resultant force components have the same signs and directions as the corresponding acceleration components.
    ∑Ft=3.0 N,∑Fn=9.0 N\sum F_t=3.0\ \mathrm{N},\qquad \sum F_n=9.0\ \mathrm{N}
Answer: The acceleration is 1.5 m/s² along the motion and 4.5 m/s² inward. Its magnitude is 4.74 m/s². The resultant force components are 3.0 N tangentially and 9.0 N inward.
Check: Each force component has units kg·m/s². If the speed increase were reduced to zero, the tangential component would vanish while the inward component would remain.

Worked example

Find speed from the inward force component

A 0.80 kg particle moves on a circular path of radius 2.0 m. At a particular instant, the resultant force component toward the centre is 6.4 N. Find its speed at that instant. The particle is moving counterclockwise; its tangential speed change is not needed.
Inward resultant at an instant
Inward resultant at an instantinward ntangent tparticle6.4 N inward

The arrow represents the inward resultant force component, not an individual contact force.

  1. Define the system and normal direction
    The system is the particle, viewed from a stationary frame. At this instant, positive normal points toward the centre of the given circular path; positive tangent follows counterclockwise motion. The force, mass, and radius are known, and the speed at this instant is unknown.
  2. Choose the governing equation
    Only the inward force component is needed to determine speed. Apply Newton’s second law in the normal direction. The tangential equation is not needed because no tangential quantity is requested.
    ∑Fn=mv2ρ\sum F_n=m\frac{v^2}{\rho}
  3. Solve and substitute
    Rearrange for the nonnegative speed, then substitute the inward force, mass, and radius. The given force component is positive under the inward-positive convention.
    v=(6.4 N)(2.0 m)0.80 kg=4.0 m/sv=\sqrt{\frac{(6.4\ \mathrm{N})(2.0\ \mathrm{m})}{0.80\ \mathrm{kg}}}=4.0\ \mathrm{m/s}
Answer: The particle’s speed is 4.0 m/s.
Check: The quantity inside the square root has units N·m/kg = m²/s², so the result is a speed. Substitution gives mv2/ρ=(0.80)(16)/2.0=6.4mv^2/\rho=(0.80)(16)/2.0=6.4 N inward.

Common mistakes and how to avoid them

Treating normal acceleration as zero because speed is constant.
Correction: Constant speed gives at=0a_t=0, but a curved path still requires inward acceleration v2/ρv^2/\rho.
Calling every force perpendicular to the path a normal force.
Correction: Normal acceleration is a motion component. A surface normal force is only one possible actual force.
Using outward as positive normal in one equation and inward in another without changing signs.
Correction: State the normal positive direction first and use it consistently when resolving every force.
Adding the tangential and normal acceleration magnitudes directly.
Correction: Because the components are perpendicular, combine them using the square root of the sum of their squares.

Lesson summary

  • At a point on a curved path, define a tangent along motion and an inward normal toward the local centre of curvature.
  • Tangential acceleration is the rate of speed change; normal acceleration is v2/ρv^2/\rho.
  • Resolve forces in the same local directions and apply Newton’s second law separately in each direction.
  • Check signs, SI units, and the limiting cases of constant speed and straight motion.

Check your understanding

Question 1

A particle moves at constant speed on a circle. Which statement is correct?
  1. Both tangential and normal acceleration are zero.
  2. Tangential acceleration is zero, while normal acceleration points inward.
  3. Normal acceleration is zero, while tangential acceleration points along the motion.
  4. Both acceleration components point along the motion.
Show answer and explanation
Tangential acceleration is zero, while normal acceleration points inward.
Constant speed means v˙=0\dot v=0. The direction changes on the circle, so v2/ρv^2/\rho remains as an inward normal component.

Question 2

A particle has speed 5.0 m/s on a path with radius of curvature 10 m. What is its normal acceleration magnitude?
  1. 0.50 m/s²
  2. 2.5 m/s²
  3. 5.0 m/s²
  4. 50 m/s²
Show answer and explanation
2.5 m/s²
Using an=v2/ρa_n=v^2/\rho, the result is (5.0 m/s)2/(10 m)=2.5 m/s2(5.0\ \mathrm{m/s})^2/(10\ \mathrm{m})=2.5\ \mathrm{m/s^2} inward.

Key terms

Tangential direction
The instantaneous direction along the path, tangent to it.
Normal direction
The direction perpendicular to the tangent; taken inward toward the local centre of curvature here.
Radius of curvature
The radius of the circle that locally matches the bend of a smooth path.
Resultant force
The vector sum of all actual forces acting on the particle.

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