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Newton’s second law for rotation
(转动定律）
应用物理系—张春华
Key words:
Particle （质点）
Rigid body（刚体）
Rotational inertia（转动惯量）
Torque（力矩）
1. The rotational variables:
REVIEW!
s

r
(1) Angular position:
（角位置）
(2) Angular displacement:
（角位移）
(3) Angular velocity
（角速度）
   f   i
Rotation axis
d

dt
(4) Angular acceleration
（角加速度）
d

dt


C
m
R
Reference line

Zero angular
position
2. Relationship of the linear and
angular variables
Distance:
Speed:
REVIEW!
s  r
v  r
Tangential acceleration:
at  r
Radial acceleration:
an   r
2
REVIEW!
3. Rotational inertia
（转动惯量）:
I   mi ri
2
axis
I   r dm
2
For rigid body :
dm
For particle:
I=mr2
M
Unit: kg.m2
The rotational inertia involves
both mass and its distribution.
r
Newton’s second law for rotation
(转动定律）
（1）Torque (力矩）
Torque is a turning or twisting
action on a body about a rotation
axis due to a force.
  rF
z
F
Ft
O
r
d
  rF sin   rFt  Fd
Where and  is the angle
between r and F , Ft is the
component of
perpendicular
F
to
.
r
Unit: N.m


P
fji
Notes:
F2
R
O r1
fij
F1
mi
mj
d
o
(1) Torque is a vector. Its direction always along the
axis (for rotating about fixed axis), either
positive or negative. (right hand rule)
(2) If several torque act on a rigid body that rotate
about a fixed axis, the net torque is the sum of
individual torque.  net  1   2   3     i
(3) The net torque of internal forces of rigid body is
zero.
  f d 0

i

i
i
(2) Newton’s second law for rotation (转动定律）
Newton’s second law：
F net  ma
For rigid body:
Treat the rigid body as a collection of particles,
for one mass element mi : F i and f i are the
external and internal forces acting on the mass
element .
Then we have:
z
ω,α
Fit
ri
O
Fi
F i  f i  mi a i


mi
fit
fi
Only the tangential
component of the
force can accelerate
the particle along the
path. So:
Fit  f it  mi ait
Fit ri  f it ri  mi ait ri  mi ri 
2
For whole rigid body:

Fit ri   fit ri   mi ri2

That is:
i
 0  I
 net  I
 net  I
 net  I
Compare with:
F net  ma
Review & summary:
 net  I
  rF
I   mi ri2
 net   i
I   r 2 dm
d

dt
Sample problem:
Fig. Shows a uniform disk, with mass M=2.5kg and
radius R=20cm, mounted on a fixed horizontal axle. A
block with mass m=1.2kg hangs from a mass-less cord
that is wrapped around the rim of the disk. Find the
acceleration of the falling block, the angular acceleration
of the disk, and the tension in the cord. The cord does
not slip and there is no friction at the axle. (I=1/2MR2)
R
Axis O
M
m
α
M ·R
T ′= - T
a
T
m
mg
Solution:
For M:
 = TR = I 
mg – T = ma
For m:
a = R
α
(1)
(2)
M· R
(3)
T
From above equations, we find the answer:
a=4.8 m/s2
T ′= - T
=24 rad/s2
T=6.0 N
Home work:
a
m
mg