Dynamics Chris Parkes October 2012 Dynamics Rotational Motion Angular velocity Angular “suvat”...

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Dynamics

Chris ParkesOctober 2012

Dynamics

Rotational Motion

Angular velocity

Angular “suvat”

Kinetic energy – rotation

Moments of Inertia

KE – linear, rotation

http://www.hep.manchester.ac.uk/u/parkes/Chris_Parkes/Teaching.html

Part III – “What we know is a drop, what we don't know is an ocean.”

READ the Textbook!

Rotation of rigid bodyabout an axis

• θ in radians– 2πr is circumference

• Angular velocity– same for all points in

body

• Angular acceleration

θ

s = rθr

v

r is perpendicular distance to axis

dθr

ds

Rotational Motion

Linear RotationalPosition, x Angle, θ

Velocity, v angular velocity, ω

Acceleration angular acceleration, α

This week – deal with components, next week with vectors

Rotational “suvat”derive:

constant angular acceleration

Rotational Kinetic Energy

O

m1

r1

m2

m3

r3

r2

ω

X-section of a rigid body rotating about an axis through O which is perpendicular to the screen

r1 is perpendicular dist of m1 from axis of rotation

r2 is perpendicular dist of m2 from axis of rotation

r3 is perpendicular dist of m3 from axis of rotation

Moment of Inertia, I• corresponding angular quantities for linear

quantities– x; v; pL– Mass also has an equivalent: moment of Inertia, I– Linear K.E.:– Rotating body v, mI:– Or p=mv becomes:Conservation of ang. mom.:e.g. frisbee solid sphere hula-hoop

pc hard disk neutron star space station

221.. mvEK

221.. IEK

IL

2211 II

R

R

R1

R2

221 MRI

252 MRI

)( 22

212

1 RRMI

masses m distance from

rotation axis r dmrrmI

iii

22

Four equal point masses , each of mass 2 kg are arranged in the xy plane as shown. They are connected by light sticks to form a rigid body. What is the moment of inertia of the system about the y-axis ?

Moment of Inertia Calculations

Systems of discrete particles

2 kg

2 kg 2 kg

2 kg1m

2m

2ii

irmI

I = 2 x (2 x12) + (2 x 22)

= 12 kg m2

y

x

Parallel-Axis Theorem

xA

yA

• Proof

Since centre-of-mass is origin of co-ordinate system

• Moment of Inertia around an axisaxis parallel to first at distance d

– Co-ordinate system origin at centre-of-mass

Table 9.2 Y& F p 291 (14th Edition)

Translation & Rotation • Combining this lecture and previous ones

• Break problem into– Velocity of centre of mass– Rotation about axis

ω

Rolling without Slipping

• Bicycle wheel along road

• Centre – pure translation

• Rim –more comlex path known as cycloid

• If no slipping

R

M

m

h

R

M

m

h

v

A light flexible cable is wound around a flywheel of mass M and radius R. The flywheel rotates with negligible friction about a stationary horizontal axis. An

object of mass m is tied to the free end of the cable. The object is released from rest at a distance h above the floor. As the object falls the cable unwinds without

slipping. Find the speed of the falling object and the angular speed of the flywheel just as the object strikes the floor.

R

M

m

h

v

A light flexible cable is wound around a flywheel of mass M and radius R. The flywheel rotates with negligible friction about a stationary horizontal axis. An

object of mass m is tied to the free end of the cable. The object is released from rest at a distance h above the floor. As the object falls the cable unwinds without

slipping. Find the speed of the falling object and the angular speed of the flywheel just as the object strikes the floor.

KEY POINTS

• Assume flywheel is a solid cylinder

• Note – cable is light – hence ignore its mass

• Note – cable unwinds without slipping - speed of point on rim of flywheel is same as that of the

cable

• If no slipping occurs, there must be friction between the cable and the flywheel

• But friction does no work because there is no But friction does no work because there is no movement between the cable and the flywheelmovement between the cable and the flywheel

Example

A bowling ball of radius R and mass M is rolling without slipping on a horizontal surface at a velocity v. It then rolls without slipping up a slope to a height h before momentarily stopping and then rolling back down. Find h.

vh

v = 0ω = 0

No slipping so no work done against No slipping so no work done against friction – Mechanical energy friction – Mechanical energy

conservedconserved