Transcript Torque, Energy, Rolling

Physics 111: Mechanics
Lecture 10
Dale Gary
NJIT Physics Department
Torque, Energy and Rolling
Torque, moment of inertia
 Newton 2nd law in rotation
 Rotational Work
 Rotational Kinetic Energy
 Rotational Energy
Conservation
 Rolling Motion of a Rigid
Object

March 26, 2016
Force vs. Torque




Forces cause accelerations
What cause angular accelerations ?
A door is free to rotate about an axis through O
There are three factors that determine the
effectiveness of the force in opening the door:



The magnitude of the force
The position of the application of the force
The angle at which the force is applied
March 26, 2016
Torque Definition
Torque, t, is the tendency of a force to rotate
an object about some axis
 Let F be a force acting on an object, and let r
be a position vector from a rotational center to
the point of application of the force, with F
perpendicular to r. The magnitude of the
torque is given by

t  rF
March 26, 2016
Torque Units and Direction
The SI units of torque are N.m
 Torque is a vector quantity
 Torque magnitude is given by

t  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
March 26, 2016
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
 St  t1 + t2  F1d1 – F2d2
 If St  0, starts rotating
 Rate of rotation of an
 If St  0, rotation rate
object does not change,
does not change
unless the object is acted

on by a net torque
March 26, 2016
General Definition of Torque




The applied force is not always perpendicular to the
position vector
The component of the force perpendicular to the
object will cause it to rotate
When the force is parallel to the position vector, no
rotation occurs
When the force is at some angle, the perpendicular
component causes the rotation
March 26, 2016
General Definition of Torque

Let F be a force acting on an object, and let r be
a position vector from a rotational center to the
point of application of the force. The magnitude
of the torque is given by
t  rF sin 
  0° or   180 °:
torque are equal to zero
   90° or   270 °: magnitude of torque attain
to the maximum

March 26, 2016
Understand sinθ
The component of the force
(F cos  ) has no tendency
to produce a rotation
 The moment arm, d, is the
perpendicular distance from
the axis of rotation to a line
drawn along the direction of
the force

t  rF sin   Fd
d = r sin
March 26, 2016
The Swinging Door

Three forces are applied to a door, as shown in
figure. Suppose a wedge is placed 1.5 m from
the hinges on the other side of the door. What
minimum force must the wedge exert so that
the force applied won’t open the door? Assume
F1 = 150 N, F2 = 300 N, F3 = 300 N, θ = 30°
F2
F3
2.0m
θ
F1
March 26, 2016
Newton’s Second Law for a
Rotating Object

When a rigid object is subject to a net torque (≠0),
it undergoes an angular acceleration
St  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
March 26, 2016
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, once to each part
 Translation:
 Rotation:


Fnet   Fi  ma



tnet   ti  I
Note: can have
Fnet = 0
but tnet ≠ 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
March 26, 2016
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 ?

March 26, 2016
Newton 2nd Law for Rotation




Draw free body
diagrams of each object
Only the cylinder is
rotating, so apply
S t = I
The bucket is falling, but
not rotating, so apply
S F = ma
Remember that a = r
and solve the resulting
equations
r
a
mg
March 26, 2016
•
•
•
•
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
F
y
 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
Mg
t 0  + Tr  I

I
Tr m( g  a)r
 1 2
Mr
I
2
1
Mr 2
2
Unknowns: a, 
So far: 2 Equations, 3 unknowns Need a constraint:
Substitute and solve:
2mgr 2m r 2


2
Mr
Mr 2


 1 + 2
m
M
 2mg

 Mr

mg
a  r
from “no
slipping”
assumption
mg
( 24 rad / s 2 )
r (m + M / 2)
March 26, 2016
•
•
•
•
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
F
y
 ma  mg  T
T  m( g  a )
mg
r
Unknowns: T, a

mg
( 24 rad/s2 )
r (m + M / 2)
a
mg
( 4.8 m/s 2 )
(m + M / 2)
support force
at axis “O” has
zero torque
a
T  m( g  a)  1.2(9.8  4.8)  6 N
mg
1
1
x f  xi  vi t + at 2  0 + (4.8 m/s 2 )(3 s)2  21.6 m
2
2
March 26, 2016
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)
March 26, 2016
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, ignoring
rotation.


Wnet   dW   F  d s
 Work


Power


dW
ds  
P
 F
 F v
dt
dt
March 26, 2016
Mechanical Energy Conservation


Energy conservation
When Wnc = 0,
Wnc  K + U
K f + U f  U i + Ki

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
March 26, 2016
Total Energy of a System
A ball is rolling down a ramp
 Described by three types of energy





Gravitational potential energy
U  Mgh
2
Translational kinetic energy K t  1 MvCM
Rotational kinetic energy
Total energy of a system
2
1 2
K r  I
2
1
1 2
2
E  MvCM + Mgh + I 
2
2
March 26, 2016
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  td
March 26, 2016
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   td   Id   I
d   Id
dt
i
i
i
i
i

I is constant for rigid object
f
f
1 2 1 2
W   Id  I  d  I f  Ii
2
2
i
i

Power
dW
d
P
t
 t
dt
dt
March 26, 2016

An motor attached to a grindstone exerts a constant torque
of 10 N-m. The moment of inertia of the grindstone is I = 2
kg-m2. The system starts from rest.

Find the kinetic energy after 8 s

Find the work done by the motor during this time
1 2
t
K f  I  f  1600 J   f  i +  t  40 rad/s     5 rad/s 2
2
I
f
W   td  t ( f   i )  10 160  1600 J
i

1
1
1
( f  i )  i t +  t 2  160 rad 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

 200 watts
dt
8
Find the instantaneous power at t = 8 s
P  t  10  40  400 watts
March 26, 2016
Work-Energy Theorem

For pure translation
Wnet  K cm  K cm , f  K cm,i 

For pure rotation
Wnet  K rot  K rot , f  K rot ,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  Ktotal  ( K rot , f + K cm, f )  ( K rot ,i + K cm,i )
1 2 1 2  1 2 1 2
  I  f + mv f    I i + mvi 
2
2
2
 2

March 26, 2016
Energy Conservation


Energy conservation
When Wnc = 0,
Wnc  Ktotal + U
K rot , f + K cm, f + U f  K rot ,i + K cm,i + U i

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 + mv 2f + mgy f
2
2
2
2

Remember, this is for conservative forces, no
dissipative forces such as friction can be present
March 26, 2016
Total Energy of a Rolling System
A ball is rolling down a ramp
 Described by three types of energy


Gravitational potential energy

U  Mgh
Translational kinetic energy


1
K t  Mv 2
2
Rotational kinetic energy
1
K r  I 2
2
Total energy of a system
1
1 2
2
E  Mv + Mgh + I
2
2
March 26, 2016
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

March 26, 2016
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 ball 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
10
v f  ( gh)1/ 2
7
March 26, 2016
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
1
12
7
1
1

mgh  mv f 2 + I  f 2  mv f 2 +  mR 2  (v f / R) 2  mv f 2
2
25
10
2
2


Box sliding
1
1
2
2
mvi + mgyi  mv f + mgy f
2
2
7
1
rolling: mgh  mv f 2
sliding: mgh  mv f 2
10
2
March 26, 2016
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.

March 26, 2016

Find the translational speeds of the blocks after block
2 descends through a distance h.
K rot , f + K cm, f + U f  K rot ,i + K cm,i + U i
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
March 26, 2016