View
4
Download
0
Category
Preview:
Citation preview
Chapter 10
Rotation(Part 3)
Outline for today
• Torque
• Newton’s Second Low for Rotation
• Work and Rotational Kinetic Energy
• Example Problems
RECAP: Rotational Inertia• Translation
• M does NOT dependon the motion!
• Rotation
• I depends on:• Shape of the body• Mass distribution• The axis of rotation
• Parallel-Axis Theorem
!
K =1
2Mv
2
!
K =1
2I" 2
!
M = mi"
!
I = Ii" = m
iri
2"
!
M = dm"
!
I = r2" dm
Parallel-Axis Theorem• If the rotational inertia (Icom) of the body about
a parallel axis that extends through the body'scenter of mass is known we can calculate therotational inertia Inew around ANY parallel axis
Inew = Icom + Mh2
where h is the perpendicular distance between thegiven axis and the axis through the center of mass
= +IcomInew
Mh2
com
Example problem• Find the rotational inertia of a thin uniform rod
with length L and mass M rotating around an axispassing through one of its end.
= +IcomInew
Mh2
com
!
Inew
= Icom
+ Mh2
!
Inew
=1
12ML
2+ M
L
2
"
# $ %
& '
2
=1
12+1
4
"
# $
%
& ' ML
2=4
12ML
2=1
3ML
2
Quiz Question• Consider a thin rotating rod with length L and
mass M. It rotates with angular velocity ω. Ifwe double the length L the rotational kineticenergy will:
A) Remain the same.B) Double.C) be half the original value.D) be four times the original value
How hard is it to open a door?
• It depends!• How hard we push (magnitude of the force)• Where we push (where the force is applied)• In what direction we push (the direction of the force)
Torque (“twist”)
• It shows the ability of a force torotate an object.
• Units: N.m
τ
!
" = rF sin#
!
" = r#F
!
" = rFt
!
r " =
r r #
r F
Torque
• Positive: counterclockwise rotation• Negative: clockwise rotation• Zero: the extension of the force passes
through the axis of rotation.
τ > 0 τ < 0 τ = 0
F
F
F
Quiz Question• The length of a bicycle pedal arm is 0.1 m and a
downward force of 200 N is applied to the pedalby a rider. What is the magnitude of the torqueabout the pedal arm’s pivot when the arm is 30o
with respect to the vertical ? A) 10 B) 50 C) 100 D) 150 E) 200
!
" = Frsin#
!
" = 200 # 0.1# sin30 =10N .m
F r
Newton’s Second Law for Rotation• Newton’s II law: The net force applied
to a body results in a acceleration so that
• What about rotation? What is the resultwhen we apply a torque?
!
r F
net= m
r a
!
" = Frsin# = (ma)rsin# = matr
" = m($r)r = mr2$
" = I$
x
y F
FtFr
φ
!
r " net
=r " i# = I
r $
Example problem• Two particles of equal mass m are attached to the
ends of a rigid, massless rod of length L1+L2. Therod is held horizontally on a fulcrum and thenreleased. What are the magnitudes of the angularacceleration of particle 1 and 2 ?
L1 L2
Fg Fg
m m
!
"net
= I#
!
" net = "1
+ "2
= Fg1L1 # Fg2L2
" net = mg(L1# L
2)
!
I = m1L1
2+ m
2L2
2= m(L
1
2+ L
2
2)!
" = #net/I
!
" = gL1# L
2
L1
2+ L
2
2
Work and Rotational Kinetic Energy
!
W = "K = K f #Ki
W =1
2mv f
2#1
2mvi
2
W =1
2mr
2$ f
2#1
2mr
2$ i
2
W =1
2I$ f
2#1
2I$ i
2
x
y
!
r v
mr
• Work
• Power
!
P =dK
dt=1
2Id
dt(" 2
) = I"d"
dt
P = I"# = $"
• Variable torque.
• Constant torque.
• Power (2)
Work and Torque
x
y FFt
Fr
φ
!
dW =r F " d
r s
dW = Ft ds = Ftr d# = $ d#
W = $# i
# f
% d#Δθ
!
W = "#$ , #$ = $ f %$i
!
P =dW
dt= "
d#
dt= "$
Example problem• A yo-yo-shaped device mounted on a horizontal frictionless axis is
used to lift a 30 kg box. The outer radius R of the device is 0.5 m andthe radius r of the hub is 0.2 m. When horizontal force Fapp=140N isapplied the box has an upward acceleration of 0.8 m/s2. What is therotational inertia of the devise about its axis of rotation?
m=30 kgR=0.5 mr=0.2 mFapp=140 Na=0.8 m/s2
_________I=?
!
" net = I# $ I =" net
#
" net = FappR %Tr
for the hanging body : T %mg = ma
T = m(g + a) and # = a /r
I =r
a(FappR % (g + a)mr)
I =0.2
0.8(140 & 0.5 % (9.8 + 0.8) & 30 & 0.2)
I =1.6 kg.m2
mg
T
T
Recommended