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UY1: Sphere On An Incline

Derive speed and acceleration for a sphere rolling down an incline using energy, force-torque, and torque-about-contact methods.

Reading settings
  • University Physics Year 1
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Enlarged diagram

Cluster role + pathway links

This page is the legacy application page for incline rolling analysis in the UY1 mechanics pathway.

1) At a glance

  • Prerequisite: Newton’s second law, torque, I_CM, energy conservation.
  • Outcomes:
    • Compute v_CM at the bottom from height drop h.
    • Derive a_CM for pure rolling on slope angle θ.
  • Find required static friction and no-slip condition.
  • Key result for a solid sphere: a_CM = (5/7)g sin θ, f = (2/7)Mg sin θ
Common traps (friction + signs)
  • Static friction is not automatically “energy loss”: in ideal pure rolling it provides torque but does no work at the instantaneous contact point.
  • Don’t assume friction always points up the slope; its direction is set by the tendency to slip at the contact.
  • Use v_CM = Rω and a_CM = Rα only for no-slip motion.

2) Setup

Model and assumptions:

  • Solid sphere, mass M, radius R, I_CM = (2/5)MR².
  • Rigid sphere and rigid incline.
  • Pure rolling: v_CM = Rω, a_CM = Rα.

Coordinates/signs:

  • Positive x down the incline.
  • Positive torque is the rolling direction down the incline.

Forces:

  • Weight component along plane: Mg sin θ (downhill).
  • Normal reaction N perpendicular to plane.
  • Static friction f up the plane (for rolling downhill).

Intuition (why friction points up the plane here):

  • Gravity pulls the CM downhill, so without friction the sphere would start sliding with little/no rotation.
  • Static friction acts up the plane to oppose that relative slipping, and its torque about the CM spins the sphere up in the rolling direction.

3) Core derivation

A) Speed at bottom from energy

If the vertical drop is h:

Mgh = (1/2)Mv_CM² + (1/2)I_CMω²

Use ω = v_CM/R:

v_CM = square root of (2gh/(1 + I_CM/MR²))

For solid sphere, I_CM/(MR²) = 2/5:

v_CM = square root of ((10/7)gh)

B) Acceleration from force + torque

Translation along slope:

Mg sin θ - f = Ma_CM

Torque about CM:

fR = I_CMα = I_CMa_CM/R

So

f = (I_CM/R²)a_CM = (2/5)Ma_CM

Substitute into translation equation:

Mg sin θ - (2/5)Ma_CM = Ma_CM

a_CM = (5/7)g sin θ

Then

f = (2/5)Ma_CM = (2/7)Mg sin θ

C) Checks

  • Units: a_CM has units of g times dimensionless factor, so m s⁻².
  • Limit θ → 0: a_CM → 0 and f → 0, consistent with level surface idealization.
  • Limit I → 0 in general formula a = (g sin θ)/(1 + I/(MR²)) gives a → g sin θ (sliding point-mass limit).

4) Worked example

Worked example 1

A solid sphere rolling down an incline

Problem

A solid sphere starts from rest and rolls without slipping down a 30° incline through a vertical drop of h = 1.4 m. Find its speed at the bottom and its acceleration.

Show full solution

  1. Find the speed by energy

    Method

    Use Mgh = (1/2)Mv² + (1/2)Iω² with I = (2/5)MR² and ω = v/R.

    Reason

    Static friction does no work in rolling, so mechanical energy is conserved.

    Working

    Answer

    Mgh = (7/10)Mv² ⇒ v = square root of ((10/7)gh) = square root of ((10/7)(9.81)(1.4)).

  2. Find the acceleration

    Method

    Use a = (g sin θ)/(1 + I/(MR²)) = (5/7)g sin θ.

    Reason

    Combining the force and torque equations gives a constant acceleration.

    Working

    Answer

    a = (5/7)(9.81)(0.5).

  3. State the answer

    Working

    Answer

    v_CM ≈ 4.43 m s⁻¹ and a_CM ≈ 3.50 m s⁻².

5) Practice set

Check your understanding 1

A body rolls without slipping down an incline of angle θ. Write a_CM in terms of I_CM.

Show hint

Combine Mg sin θ-f = Ma with fR = Iα and a = Rα.

Show answer

From Mg sin θ-f = Ma and fR = I_CMa/R: f = I_CMa/R², so a_CM = (g sin θ)/(1 + I_CM/(MR²)).

Check your understanding 2

For a solid sphere rolling down a 20° incline, find the friction force as a fraction of its weight.

Show hint

Use f = (2/7)Mg sin θ.

Show answer

f = (2/7)Mg sin θ, so f/(Mg) = 2/7 sin 20° ≈ 0.0977.

Check your understanding 3

What condition on μₛ prevents a solid sphere from slipping on the incline?

Show hint

N = Mg cos θ.

Show answer

f ≤ μₛN with f = (2/7)Mg sin θ and N = Mg cos θ gives 2/7 tan θ ≤ μₛ, that is tan θ ≤ (7/2)μₛ.

6) Summary and next steps

For a sphere on an incline, rotational inertia reduces linear acceleration below g sin θ. For a solid sphere:

a_CM = (5/7)g sin θ, f = (2/7)Mg sin θ, v_bottom = square root of ((10/7)gh)

Next steps:

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Syllabus and review details

No official syllabus alignment is listed for this lesson.

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