11 Dynamics Rotation 1

Halliday, Resnick, Walker: Chapter 10

Recall:Kinematics deals with the concepts/terminology needed to describe motion.

We have been discussing rotational kinematics.We have been discussing rotational kinematics.

Dynamics deals with the effect that forces have on motion.

Rotational Dynamics IRotational Dynamics I

We now discuss rotational dynamics.We now discuss rotational dynamics.

2

Now we will discuss the forces that give rise to motion (and the resulting motion).

We now discuss rotational dynamics.We now discuss rotational dynamics.

Start by considering pure rotational motion pure rotational motion about a fixed axis i.e. an axis that does not move.


Kinetic Energy of RotationConsider an object rotating about a fixed axis.

Every point on the object travels in a circle in the plane perpendicular to the axis of rotation.

[Halliday, Resnick, Walker -- 10.6: Kinetic Energy of Rotation]

K = L+++ 2332

12222

12112

1 vmvmvm

∑2

21

iivmK =

Recall: vi = ri ω �( ) ( ) 22

212

212

21 ωω ∑∑∑ === iiiiii rmrmvmK

Consider the object’s kinetic energy:

Rotational Dynamics IRotational Dynamics I[Halliday, Resnick, Walker -- 10.6: Kinetic Energy of Rotation]

( ) 22

21

2

21 ω∑=∑=

iiiirmvmK

∑= 2iirmI

definition: moment of inertia, I

Kinetic Energy of RotationConsider an object rotating about a fixed axis.

∑ ii

2

21 ωIK =

For a body rotating about a fixed axis:

* This is shown for completeness; you will not be responsible for this.

For continuous mass distributions:*

∫= dmrI 2

Example:A rigid body consists of two particles of mass m connected by a rod of length L and negligible mass.


(a) Determine the moment of inertia about an axis through the center of mass, perpendicular to the rod.

(b) Determine the moment of inertia about an axis through the left end, perpendicular to the rod.

Example:Four masses (of mass m0) sit at the corners of a square (having sides of length L).


Determine the moment of inertia about an axis through the center of the square, perpendicular to its plane (i.e. an axis through point O in the figure).


The force can be resolved into two components –

Consider a force F acting on a point mass constrained to move in a circle.

Torque and Angular Acceleration[Halliday, Resnick, Walker – 10.8, 9: Newton’s Second Law for Rotation]

resolved into two components –(1) the radial component Fr=|F| cos φ(2) the tangential component Ft=|F| sin φ

Ft gives rise to tangential acceleration; the tangential acceleration is related to the angular acceleration: at = α r


Torque and Angular Acceleration

at = α r �� Ft = m α rNewton’s 2nd Law: Ft = mat

Multiply both sides by r �rFt = m α r2

[Halliday, Resnick, Walker – 10.8, 9: Newton’s Second Law for Rotation]

definition: torque, ττ = r Ft = r |F| sin φ

Notice: I = mr2 – moment of inertia for a particle moving in a circle

Result: τ = I αTorque gives rise to angular acceleration.


Torque and Angular AccelerationResult: τ = I αwhere τ = r |F| sinφ and I=mr2

Note: (1) We identifycounterclockwise torques as positive


counterclockwise torques as positive and clockwise torques as negative.

(2) The unit of torque* is Newton x meter.

(3) α must be expressed in radians/s2.

*This is the same as the unit for energy, however the quantities are not related.


Recall: Angular velocity and angular acceleration are vectorsvectors.

The directionassociated with these quantities is parallel to the axis of rotation – the direction is determined by the “right hand


direction is determined by the “right hand rule”.

Example: angular velocityThe fingers curl around the record and point in the direction it is moving; the extended thumb points in the direction of ωωωω.


whereτ = r Ft = r |F| sin φWe haveτ = I α

Notice: If we write ττττ = r x F

(1) |ττττ| = r |F| sin φ


Note: This is consistent with the definition of angular acceleration.

(1) |ττττ| = r |F| sin φ(2) direction – perpendicular to F and r

Result: ττττ = I αααα where ττττ = r x F


We considered a force F acting on a point mass constrained to move in a circle.


Result: ττττ = I ααααwhere ττττ = r x F and I = mr2


The internal forces and, hence, internal torques cancel due to Newton’s 3rd law � we need only consider the external torques.

Break the body up into a collection of point masses.

Now consider a rigid body rotating about a fixed axis:


Torque and Angular AccelerationNow consider a rigid body rotating about a fixed axis:

We need only consider the external torques�

Σiττττi = Σi mi ri2αααα


where ττττi = ri x Fi and I = Σi mi ri2

Σiττττi = Σi mi ri αααα

Result: Σiττττi = Iαααα

sum of external torques on a rigid body = moment of inertia x angular acceleration



Example: opening a door

Result: Σiττττi = IααααThe angular acceleration is largest

when the (net) torque –ττττ = r x F – is largest.


Example: opening a doorF || r � ττττ = 0 –F does nothing toward opening the door

|ττττ |= |r||F|sinφ – The most effective way to open the door is to push perpendicular to the door and far from the hinge.

Example:Compute the torque (magnitude and direction) about point O due to the force F – |F|=16.0N; the rod has length 4.00m


(a)

(b)(b)

(c)

Rotational Dynamics IRotational Dynamics I[Halliday, Resnick, Walker -- 10.6: Kinetic Energy of Rotation]

2212

21 mvKIK == ω

Note the similarity between the rotational and linear equations.

Comparison: Translation and RotationTranslation Rotation

F = ma ττττ = Iαααα

In particular, moment of inertia is analogous to mass –

mass = measure of translational inertia of an objectmoment of inertia = measure of rotational inertia of an object

relative to some fixed axis of rotation

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11 Dynamics Rotation 1