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Newton’s Second Law for Rotation (AQA A Level Physics) : Revision Note

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Newton’s Second Law for Rotation

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

F space equals space m a

  • 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

tau space equals space I alpha

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

    • tau = torque (N m)

    • I = moment of inertia (kg m2)

    • alpha = 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 space equals space m a

Torque:  tau space equals space F r

  • Where:

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

  • Combining these equations gives:

tau space equals space r open parentheses m a close parentheses

  • The moment of inertia of a rotating body can be thought of as analogous to (the same as) 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:  alpha space equals space a over r

Moment of inertia (point mass):  I space equals space m r squared

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

tau space equals space r open parentheses m r alpha close parentheses

tau space equals space open parentheses m r squared close parentheses alpha

tau space equals space I alpha

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

Linear variable

Rotational variable

Force, F

Torque, tau

Mass, m

Moment of inertia, I

Acceleration, a

Angular acceleration, alpha

Newton's Second Law, F space proportional to space a

Newton's Second Law, tau space proportional to space alpha

F space equals space m a

tau space equals space I alpha

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 1 half M R squared.

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 space equals space m a

m g space minus space T space equals space m a  eq. (1)

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

tau space equals space I alpha

T R space equals space I alpha

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

  • The angular acceleration alpha of the pulley is: 

alpha space equals space a over R

  • Substitute this into the previous equation:

T R space equals space I a over R

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

Moment of inertia of the cylinder:  I space equals space 1 half M R squared

T R space equals space open parentheses 1 half M R squared close parentheses a over R

T space equals space open parentheses 1 half M R squared close parentheses a over R squared space equals space 1 half M a  

T space equals space 1 half M a  eq. (2)

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

m g space minus space 1 half M a space equals space m a

m g space equals space m a space plus space 1 half M a space equals space a open parentheses m space plus space M over 2 close parentheses

Acceleration of the block:  a space equals space fraction numerator m g over denominator m space plus space M over 2 end fraction

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Ashika

Author: Ashika

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