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Transcript
Physics 106: Mechanics
Lecture 04
Wenda Cao
NJIT Physics Department
Rotational Work and Energy
Torque, moment of inertia
 Newton 2nd law in rotation
 Rotational Work
 Rotational Kinetic Energy
 Rotational Energy
Conservation
 Rolling Motion of a Rigid
Object

February 10, 2009
Torque Units and Direction
The SI units of torque are N.m
 Torque is a vector quantity
 Torque magnitude is given by

  rF sin   Fd

Torque will have direction


If the turning tendency of the force is counterclockwise,
the torque will be positive
If the turning tendency is clockwise, the torque will be
negative
February 10, 2009
Net Torque
The force F1 will tend to
cause a counterclockwise
rotation about O
 The force F2 will tend to
cause a clockwise
rotation about O
 S  1 + 2  F1d1 – F2d2
 If S  0, starts rotating
 Rate of rotation of an
 If S  0, rotation rate
object does not change,
does not change
unless the object is acted

on by a net torque
February 10, 2009
Newton’s Second Law for a
Rotating Object

When a rigid object is subject to a net torque (≠0),
it undergoes an angular acceleration
S  I
The angular acceleration is directly proportional to
the net torque
 The angular acceleration is inversely proportional to
the moment of inertia of the object
 The relationship is analogous to

 F  ma
February 10, 2009
Strategy to use the Newton 2nd Law
Draw or sketch system. Adopt coordinates, indicate rotation
axes, list the known and unknown quantities, …
• Draw free body diagrams of key parts. Show forces at their
points of application. find torques about a (common) axis
•
• May need to apply Second Law twice to each part


 Translation: Fnet   Fi  ma



 Rotation:
net   i  I
Note: can have
Fnet .eq. 0
but net .ne. 0
• Make sure there are enough (N) equations; there may be
constraint equations (extra conditions connecting unknowns)
• Simplify and solve the set of (simultaneous) equations.
• Find unknown quantities and check answers
February 10, 2009
The Falling Object
A solid, frictionless cylindrical reel of
mass M = 2.5 kg and radius R = 0.2 m
is used to draw water from a well. A
bucket of mass m = 1.2 kg is attached
to a cord that is wrapped around the
cylinder.
 (a) Find the tension T in the cord and
acceleration a of the object.
 (b) If the object starts from rest at the
top of the well and falls for 3.0 s before
hitting the water, how far does it fall ?

February 10, 2009
Newton 2nd Law for Rotation




Draw free body
diagrams of each object
Only the cylinder is
rotating, so apply
S=I
The bucket is falling, but
not rotating, so apply
SF=ma
Remember that a =  r
and solve the resulting
equations
r
a
mg
February 10, 2009
•
•
•
•
Cord wrapped around disk, hanging weight
Cord does not slip or stretch  constraint
Disk’s rotational inertia slows accelerations
Let m = 1.2 kg, M = 2.5 kg, r =0.2 m
For mass m:
y
T
mg
Fy  ma  mg  T
T  m (g  a)
r
Unknowns: T, a
support force
at axis “O” has
zero torque
a
FBD for disk, with axis at “o”:
N
T
  0  + Tr  I

Mg
I
Tr m(g  a)r
 1
I
Mr 2
1 2
Mr
2
Unknowns: a, 
2
So far: 2 Equations, 3 unknowns Need a constraint:
Substitute and solve:
2mgr 2mr 2

2
Mr
Mr 2
mg
(1 + 2
m
2mg
)
M
Mr

a  + r
from “no
slipping”
assumption
mg
( 24 rad/s 2 )
r(m + M/2)
February 10, 2009
•
•
•
•
Cord wrapped around disk, hanging weight
Cord does not slip or stretch  constraint
Disk’s rotational inertia slows accelerations
Let m = 1.2 kg, M = 2.5 kg, r =0.2 m
For mass m:
y
T
mg
Fy  ma  mg  T
T  m (g  a)
r
Unknowns: T, a

mg
( 24 rad/s 2 )
r(m + M/2)
a
mg
( 4.8 m/s 2 )
(m + M/2)
T  m ( g  a)  1.2(9.8 - 4.8)  6N
support force
at axis “O” has
zero torque
a
mg
1
1
x f - x f  vi t + at 2  0 +  4.8  32  21.6m
2
2
February 10, 2009
Rotational Kinetic Energy
An object rotating about z axis with an angular
speed, ω, has rotational kinetic energy
 The total rotational kinetic energy of the rigid
object is the sum of the energies of all its
particles

1
K R   K i   mi ri 2 2
i
i 2
1
1 2
2 2
K R    mi ri   I
2 i
2



Where I is called the moment of inertia
Unit of rotational kinetic energy is Joule (J)
February 10, 2009
Work-Energy Theorem for pure
Translational motion

The work-energy theorem tells us
Wnet
1 2 1 2
 KE  KE f  KEi  mv f  mvi
2
2
Kinetic energy is for point mass only, no rotation
 Work



Wnet   dW   F  d s

Power


dW
ds  
P
 F
 F v
dt
dt
February 10, 2009
Mechanical Energy Conservation


Energy conservation
When Wnc = 0,
Wnc  KE + PE
KE f + PE f  PEi + KEi

The total mechanical energy is conserved and remains
the same at all times
1 2
1 2
mvi + mgyi  mv f + mgy f
2
2

Remember, this is for conservative forces, no
dissipative forces such as friction can be present
February 10, 2009
Total Energy of a System
A ball is rolling down a ramp
 Described by three types of energy



Gravitational potential energy

PE  Mgh
Translational kinetic energy

Rotational kinetic energy
Total energy of a system
1
Mv 2
2
1 2
KEr  I
2
KEt 
1
1 2
2
E  Mv + Mgh + I
2
2
February 10, 2009
Work done by a pure rotation


Apply force F to mass at point r,
causing rotation-only about axis
Find the work done by F applied to
the object at P as it rotates through
an infinitesimal distance ds


dW  F  d s  F cos(90   )ds
 F sin ds  Fr sin d

Only transverse component of F
does work – the same component
that contributes to torque
dW  d
February 10, 2009
Work-Kinetic Theorem pure rotation

As object rotates from i to f , work done by the
torque
f
f
f
f
f
d
W   dW   d   Id   I
d   Id
dt
i
i
i
i
i

I is constant for rigid object
f
f
i
i
W   Id  I  d 

Power
1 2 1 2
I f  Ii
2
2
dW
d
P

 
dt
dt
February 10, 2009

An motor attached to a grindstone exerts a constant torque
of 10 Nm. The moment of inertia of the grindstone is I = 2
kgm2. The system starts from rest.

Find the kinetic energy after 8 s

Find the work done by the motor during this time
1 2

K f  I f  1600 J   f  i + t  40rad/s     5rad/s 2
2
I
f
W   d   ( f   i )  10 160  1600 J
i

1
1
1
( f   i )  i t + t 2  160rad W  K f  K i  I 2f  Ii2  1600 J
2
2
2
Find the average power delivered by the motor
Pavg 

dW 1600

 200Watt
dt
8
Find the instantaneous power at t = 8 s
P    10  40  400Watt
February 10, 2009
Work-Energy Theorem

For pure translation
Wnet  KEcm  KEcm , f  KEcm ,i 

For pure rotation
Wnet  KErot  KErot, f  KErot,i 

1 2 1 2
mv f  mvi
2
2
1 2 1 2
I f  Ii
2
2
Rolling: pure rotation + pure translation
Wnet  KEtotal  ( KErot, f + KEcm, f )  ( KErot,i + KEcm,i )
1 2 1 2
1 2 1 2
 ( I f + mv f )  ( Ii + mvi )
2
2
2
2
February 10, 2009
Energy Conservation


Energy conservation
When Wnc = 0,
Wnc  KEtotal + PE
KErot, f + KEcm , f + PE f  KErot,i + KEcm ,i + PEi

The total mechanical energy is conserved and remains
the same at all times
1 2 1 2
1
1
Ii + mvi + mgyi  I 2f + mv2f + mgy f
2
2
2
2

Remember, this is for conservative forces, no
dissipative forces such as friction can be present
February 10, 2009
Total Energy of a Rolling System
A ball is rolling down a ramp
 Described by three types of energy



Gravitational potential energy

PE  Mgh
Translational kinetic energy

Rotational kinetic energy
Total energy of a system
1
Mv 2
2
1 2
KEr  I
2
KEt 
1
1 2
2
E  Mv + Mgh + I
2
2
February 10, 2009
Problem Solving Hints

Choose two points of interest


One where all the necessary information is given
The other where information is desired
Identify the conservative and non-conservative
forces
 Write the general equation for the Work-Energy
theorem if there are non-conservative forces


Use Conservation of Energy if there are no nonconservative forces
Use v = r to combine terms
 Solve for the unknown

February 10, 2009
A Ball Rolling Down an Incline

A ball of mass M and radius R starts from rest at a
height of h and rolls down a 30 slope, what is the
linear speed of the bass when it leaves the incline?
Assume that the ball rolls without slipping.
1 2
1 2 1
1
2
2
mvi + mgyi + Ii  mv f + mgy f + I f
2
2
2
2
1
1
2
2
0 + Mgh + 0  Mv f + 0 + I f
2
2
vf
2
2
I  MR  f 
5
R
2
v
1
12
1
1
2
2
2
f
Mgh  Mv f +
MR 2 2  Mv f + Mv f
2
25
R
2
5
vf  (
10 1/ 2
gh)
7
February 10, 2009
Rotational Work and Energy


A ball rolls without slipping down incline A,
starting from rest. At the same time, a box
starts from rest and slides down incline B,
which is identical to incline A except that it
is frictionless. Which arrives at the bottom
first?
Ball rolling:
1
1
1
1
2
2
2
2
mvi + mgyi + Ii  mv f + mgy f + I f
2
2
2
2
mgh 

Box sliding
1
1
2
2
mv f + I f
2
2
1
1
2
2
mvi + mgyi  mv f + mgy f
2
2
1
2
mgh  mv f
2
February 10, 2009
Blocks and Pulley

Two blocks having different masses m1 and
m2 are connected by a string passing over a
pulley. The pulley has a radius R and
moment of inertia I about its axis of
rotation. The string does not slip on the
pulley, and the system is released from rest.
Find the translational speeds of the blocks
after block 2 descends through a distance h.
 Find the angular speed of the pulley at that
time.

February 10, 2009

Find the translational speeds of the blocks after block
2 descends through a distance h.
KErot, f + KEcm , f + PE f  KErot,i + KEcm ,i + PEi
1
1
1
2
2
2
( m1v f + m2v f + I f ) + (m1 gh  m2 gh)  0 + 0 + 0
2
2
2
1
I
(m1 + m2 + 2 )v 2f  m2 gh  m1 gh
2
R
 2(m2  m1 ) gh 
vf  
2
m
+
m
+
I
/
R
2
 1


1/ 2
Find the angular speed of the pulley at that time.
1  2(m2  m1 ) gh 
f   

R R  m1 + m2 + I / R 2 
vf
1/ 2
February 10, 2009