Transcript rotary motion - GEOCITIES.ws

MOTION ABOUT AN
INTERNAL AXIS
Quantity
Type of motion
\\ linear //
Translational
(Symbol &Units)
Inertia
Displacement
m (Kg)
Δd (m)
\\ angular //
Rotational
(Symbol & Units)
I <iota>
(Kg . m2)
Δθ <theta> (Radians)
v (m/s)
ω <omega> (Rad/s)
Acceleration
a (m/s2)

Cause of
Acceleration
F (N)

Velocity
<alpha> (Rad/s2)
<tau> (m . N)
{torque}
Radian
►That
angle which, if its vertex were
placed at the center of a circle,
would subtend an arc of that circle
equal to its radius.
R
R
{Note:1rev  2Rad }
1 Rad
R
Rotary Kinematics Equations
Linear
d
v
t
v
a
t
Rotary


t
d  v0 t  at


t
v1  v0  at
   0 t  t
 1   0  t
v1  v  2ad
 1    2
1
2
2
0
2
1
2
2
0
2
Rotary-Linear Relationships
dT  R
Δt
Δ
Δt
vT ΔR
Δt
Δt
aT  R
R
Δθ
ΔdT
R
Angular Velocity Direction
Determined by “The Right-Hand Rule”
Curl the fingers of your right
hand in the direction of the
rotation and extend your
ω
thumb along the axis.
Your thumb points in the
direction of the
angular velocity.
Vt
ω
Vt
Clockwise rotation >>>>> ω is away from you.
Counter-clockwise rotation >>>>> ω is toward you.
PROBLEM
A Ferris wheel having a radius of 12 m starts at
rest and completes one revolution in 8.0 s.
Find : a) its angular acceleration,
b) its angular velocity at 8.0 s, and
c) the tangential velocity of a rider at 8.0 s.
Given:
R=12 m
VT0=0 m/s
ω0=0 rad/s
 = 1.00 rev = 2Rad
Δt=8.0 S
R
vT1
ω1
(a)
 ?
PROBLEM 0
2
1
   0 t  2 t
2 
2(2rad )
 2 
2
(8.0s)
t
  0.20rad / s
2
b)
ω1=?
PROBLEM
0
 1   0  t
2 
1 

t
2
t
( )
2(2rad )
1 
8.0s
 1  16
. rad / s
2

t
PROBLEM
(C)
VT1 = ?
VT1 = Rω1
2 R
VT1 
t
2(12m)(2rad )
VT1 
8.0s
VT1 = 19 m/s
Parallel Forces act in the same or
opposite directions at different points
on a body.
Center of gravity - That point at
which all of the weight of a body can
be considered to be concentrated.
Finding the Center of Gravity
Uniform body – find the geometric center.
C of G
C of G
Finding the Center of Gravity
For a non-uniform body - the C of G lies
directly above or below any stable point
of support.
C of G

Torque,
-The product of a force
and its perpendicular distance from the
pivot point (fulcrum). ………………………
  Fd 
Units :
m.N (meter.newtons)
Torque arm, d  -The perpendicular…
distance between the fulcrum and the line
of force.……………………………………………...
F
d
  Fd 
F
d
θ
θ
  Fd 
  Fd sin
d
d   d sin 
Also correct – and in keeping with
what the text has been promoting…
  F d
F
F
θ
d
F  F sin 
  Fd sin
1st Condition of Equilibrium
(Translational motion)
► If
there is no change in a body’s
translational motion then there is no net
force acting on it.

 at Eq ,  F  0
  Fleft   Fright
etc
2nd Condition of Equilibrium
(Rotational motion)
► If
there is no change in a body’s rotational
motion then there is no net torque acting on
it.
 at Eq,   0
   clockwise    counterclockwise
SEE-SAW TYPE PROBLEM
A 400.0 N boy sits 1.50 m from the
fulcrum of a see-saw. How far from
the fulcrum would a 300.0 N girl have
to sit in order to balance the see-saw?
Given :
Illustration :
d2
N d1
Fw1 = 400.0 N
Fw2 = 300.0 N
d1 = 1.50 m
d2 = ?
Fw2
Fw1
SEE-SAW TYPE PROBLEM
Free body diagram
Fw2
1st cond of eq :

N
d2
d1

cc
F
up
c
Fw1
  Fdown
N = Fw1 + Fw2
No help (in this problem)
2nd
SEE-SAW TYPE PROBLEM
cond of eq :   clockwise    counterclockwise
Fw1d1 = Fw2d2
Fw1d1
d2 
Fw 2
(400.0 N )(150
. m)
d2 
300.0 N
d2 = 2.00 m
SEE-SAW TYPE PROBLEM
The same problem done
using a
Force -Torque table

Fw2
F
d
- Fw1
+ d1
- Fw1d1
+N
0
0
- Fw2
- d2
+ Fw2d2

At eq:Σ F =0 and Σ =0
d2

cc
N
d1

c
Fw1

F
d
- Fw1
+ d1
- Fw1d1
+N
0
0
- Fw2
- d2
+ Fw2d2
Σ F =0
- Fw1 + N – Fw2 = 0
N = Fw1 + Fw2

Σ =0
- Fw1d1 + Fw2d2 = 0
Fw1d1 = Fw2d2
NOTE : These are the same EQs as before.
BRIDGE PROBLEMS
Ne
Nw
Fwg
C Fwb
A 650 N graduate stands 2.0 m from the west end
of a uniform 1250 N bridge that is 5.0 m long.
How much weight is supported by the supports on
each side of the bridge?
{As there is no obvious pivot point, you may
place your fulcrum anywhere on the bridge.}
{BUT placing the fulcrum at the position of one of the
unknown forces will usually simplify the algebra.}
BRIDGE PROBLEMS
Given :
Fwg = 650 N
Fwb = 1250 N
dg = 2.0 m
L = 5.0 m
BRIDGE PROBLEMS
Nw
dg L/2
Free-body diagram
Fwg
Force-torque table
F
d
0

L
Ne
Fwb
+ Nw
0
- Fwg
+ dg
- Fwgdg
- Fwb
+ L/2
- ½ FwbL
+ Ne
+L
+ NeL
{- —> Tc}
{+ —> Tcc}
BRIDGE PROBLEMS
At EQ :
ΣF=0
+ Nw – Fwg – Fwb + Ne = 0
At EQ :
  0
- Fwgdg - ½ FwbL + NeL = 0
(Note : 2 unknowns)
(Note : 1 unknown)
NeL = Fwgdg + ½ FwbL
Ne = (Fwgdg)/L + ½ Fwb
Ne = (650 N)(2.0 m)/5.0 m + ½ (1250 N)
Ne = 880 N
BRIDGE PROBLEMS
Using our two equations:
+ Nw – Fwg – Fwb + Ne = 0
Ne = (Fwgdg)/L + ½ Fwb
+ Nw – Fwg – Fwb + {(Fwgdg)/L + ½ Fwb} = 0
Nw = Fwg + Fwb - (Fwgdg)/L - ½ Fwb
Combine like terms
Nw = Fwg (1- dg/L) + ½ Fwb
Nw = 650 N (1 - 2.0 m/5.0 m) + ½ (1250 N)
Nw = 1020 N
ROTATIONAL INERTIA, I
-- The resistance of a rotating body to
any change in its angular velocity.
Units: Kg.m2
Angular inertia is a function of the
distribution of the mass over the radius.
R
R
R
r
ω
Thin ring
I = mR2
ω
Disk
I = ½ mR2
ω
Thick ring
I = ½ m (R2+ r2)
Angular inertia is a function of the
distribution of the mass over the radius.
R
ω
Disk
I= ¼
mR2
R
R
ω
Solid ball
I = 2/5 mR2
ω
Hollow ball
I = 2/3 mR2
Angular inertia is a function of the
distribution of the mass over the radius.
L
L
ω
Thin rod
I = 1/12 mL2
Thin rod
I = 1/3 mL2
ω
Newtons law of acceleration
for linear motion :
F = ma
The rotary version of Newtons
law of acceleration :
 
PROBLEM
A tangential force of 5.0 N is applied for
10.0 s to the rim of a 1.5 Kg bicycle wheel
having a radius of 25 cm.
If the wheel starts from rest, find :
(a) The rotational inertia of the wheel,
(b) Its angular acceleration,
(c) Its angular velocity at 10.0 s,
(d) The distance traveled by a point on
the rim in that 10.0 s.
PROBLEM
FT =5.0 N
FT
Δt = 10.0 s
R
ω
m = 1.5 kg
R = 25 cm = 0.25 m
ωo = 0 rad/s
a)
I (ring) = mR2
I (ring) = (1.5 kg)(0.25 m)2
I (ring) = 0.094 kg.m2
PROBLEM
b)



FT R

2
mR
5.0 N

(15
. kg )(0.25m)
FT

mR
= 13 Rad/s2
PROBLEM
0
c)
 1   0  t
FT t
1 
mR
(5.0 N )(10.0s)
1 
(15
. kg )(0.25m)
ω1 = 130 rad/s
d)
Δd = RΔθ
PROBLEM
0
   0 t  t
1
2
2
2
d  R( t )
2
FT
FT t
2
1
d  R[ 2 ( ) t ] 
2m
mR
1
2
(5.0 N )(10.0s)
d 
2(15
. kg )
2
= 170 m
On-your-own
A 7.5 kg bowling ball (Irolling = 7/5 m R2 )
with a radius of 12 cm is at the top of a
ramp 5.0 m long and inclined at 30.00.
a) Find the torque acting on the ball.
b) Find its rotational inertia.
c) Find its angular acceleration as it rolls down the ramp.
d) Find its linear acceleration as it rolls down the ramp.
e) Find its linear speed when it reaches the bottom.
f) If a hollow ball (Irolling = 5/3 m R2) with the same
mass and radius was released next to it, which
would win the race?
R
(Note that the
axis of rotation
is NOT at the
center of the
ball so I = 2/5 mR2 )
θ
R
Fw
θ