Newton’s Second Law for Rotation (DP IB Physics: HL): Revision Note

Katie M

Written by: Katie M

Reviewed by: Caroline Carroll

Updated on

Newton’s Second Law for Rotation

  • In linear motion, the force required to give an object a certain acceleration depends on its mass

F = ma

  • This is Newton's Second Law of linear motion, where:

    • F = force (N)

    • m = mass (kg)

    • a = linear acceleration (m s−2)

  • In rotational motion, the torque required to give a rotating object a certain angular acceleration depends on its moment of inertia

τ = Iα

  • This is Newton's Second Law of rotational motion, where:

    • τ = torque (N m)

    • I = moment of inertia (kg m2)

    • α = angular acceleration (rad s−2)

rZNng5oC_1-4-6-newtons-second-law-for-rotation

Newton's second law for rotating bodies is equivalent to Newton's second law for linear motion

  • This equation comes from the fact that torque is the rotational equivalent of force:

Force:  F = ma

Torque:  τ = Fr

  • Where:

    • r = distance from the axis of rotation (m)

  • Combining these equations gives:

τ = r(ma)

  • The moment of inertia of a rotating body can be thought of as analogous to mass

    • The inertia of a mass describes its ability to resist changes to linear motion, which is referring to linear acceleration

    • Similarly, the moment of inertia of a mass describes its ability to resist changes to rotational motion, which is referring to angular acceleration

Angular acceleration:  α = ar

Moment of inertia (point mass):  I = mr2

  • Using these equations with the equations for force and torque leads to:

τ = r(mrα)

τ = (mr2)α

τ = Iα

Comparison of linear and rotational variables in Newton's Second Law

Linear variable

Rotational variable

Force, F

Torque, τ

Mass, m

Moment of inertia, I

Acceleration, a

Angular acceleration, α

Newton's Second Law, F  a

Newton's Second Law, τ  α

F = ma

τ = Iα

Worked Example

A block of mass m is attached to a string that is wrapped around a cylindrical pulley of mass M and radius R, as shown in the diagram.

The moment of inertia of the cylindrical pulley about its axis is 12MR2.

YAsQNsP5_1-4-6-newtons-second-law-for-rotation-worked-example

When the block is released, the pulley begins to turn as the block falls.

Write an expression for the acceleration of the block.

Answer:

Step 1: Identify the forces acting on the block

Q2Dp7PX~_1-4-6-newtons-second-law-for-rotation-worked-example-ma

Step 2: Apply Newton's second law to the motion of the block

F = ma

mg  T = ma  eq. (1)

Step 3: Apply Newton's second law to the rotation of the pulley

τ = Iα

TR = Iα

Step 4: Write the equation for the pulley in terms of acceleration a

  • The angular acceleration α of the pulley is: 

α = aR

  • Substitute this into the previous equation:

TR = IaR

  • Substitute in the expression for the moment of inertia and simplify:

Moment of inertia of the cylinder:  I = 12MR2

TR = (12MR2)aR

T = (12MR2)aR2 = 12Ma  

T = 12Ma  eq. (2)

Step 5: Substitute eq. (2) into eq. (1) and rearrange for acceleration a

mg  12Ma = ma

mg = ma + 12Ma = a(m + M2)

Acceleration of the block:  a = mgm + M2

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Katie M

Author: Katie M

Expertise: Curriculum Expert

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.

Caroline Carroll

Reviewer: Caroline Carroll

Expertise: Head of Content Delivery

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about delivering high-quality resources to help students achieve their full potential.