m1
m2
m1=100 kg adult, m2=10 kg baby.
The seesaw starts from rest. Which direction will it rotates?
(a)
(b)
( )
(c)
(d)
Counter-Clockwise
Clockwise
N rotation
No
t ti
Not enough information
Effect of a Constant Net Torque
2.3 A constant non-zero net torque is exerted on a wheel. Which of the
following quantities must be changing?
1.
2.
3.
4.
5.
6.
angular position
angular velocity
angular acceleration
moment of inertia
kinetic energy
the mass center location
A.
B
B.
C.
D.
E.
1, 2, 3
4 5,
4,
5 6
1,2, 5
1, 2, 3, 4
2, 3, 5
1
Example: second law for rotation
PP10601-50: A torque of 32.0 N·m on a certain wheel causes an angular acceleration
of 25.0 rad/s2. What is the wheel's rotational inertia?
Second Law example: α for an unbalanced bar
Bar is massless and originally horizontal
Rotation axis at fulcrum point
+y
Î N has zero torque
Find angular acceleration of bar and the linear
acceleration of m1 just after you let go
Use:
τ net = Itot α ⇒ α =
τ net
Itot
where: Itot = I1 + I2 = m1L21 + m2L22
τ net =
∑ τ o,i = + m1gL1 − m2gL 2
What happened to sin(θ) in moment arm?
m1g
N
L2
fulcrum
m2g
Constraints:
Using specific numbers:
Let m1 = m2= m
L1=20 cm, L2 = 80 cm
α=
gL1 − gL 2
=
g(0.2 - 0.8)
+
0.2 2 + 0.8 2
2
= − 8.65 rad/s
Clockwise
net
torque
α=
total I
about
pivot
L1
m1gL1 − m2gL 2
m1L21 + m2L22
L21
L22
a1 = -α L1 = + 1.7 m/s 2
Accelerates UP
What if bar is not horizontal?
2
See Saw
‰ 3.1. Suppose everything is as it was in the preceding example, but the bar is
NOT horizontal. Assume both masses are equal. Which of the following is the
correct equation for the angular acceleration?
A))
α=
(L1 − L 2 )
L21 + L22
L1
g
m1g
fulcrum
(L1 − L 2 )
B)
g
α= 2
L1 + L22 cos( θ)
C)
α=
D))
α=
E)
α=
(L1 − L 2 )
L21 + L22
(L1 − L 2 )
L21 + L22
θ
N
L2
m2g
g sin( θ)
g cos(( θ)
(L1 − L 2 ) g
L21 + L22 sin( θ)
τ net = Itot α
See Saw - Solution
‰ 3.1. Suppose everything is as it was in the preceding example, but the bar is
NOT horizontal. Assume both masses are equal. Which of the following is the
correct equation for the angular acceleration?
L1
α = τnet / Itot
τ net =
m1g
N
fulcrum
θ
L2
∑ τ o,i = + m1gL1 cos( θ) − m2gL 2 cos( θ)
m2g
Itot = I1 + I2 = m1L21 + m2L22
α=
[m1L1 − m2L 2 ]
m1L21 + m2L22
g cos( θ)
For equal masses m1 – m2 = m
∴ α=
(L1 − L 2 )
L21
+
L22
gcos( θ)
If bar is balanced when it’s
horizontal, does that change
when it is not horizontal?
3
Method for solving (complex) problems using the Second Law
Many components in the system means several (N) unknowns….
… need an equal number of independent equations
Method:
• Draw or sketch system. Adopt coordinates, name the variables, indicate
rotation axes, list the known and unknown quantities, …
• Draw
D
free
f
body
b d diagrams
di
of
f key
k
parts. Show
Sh
forces
f
at their
h i points
i
of
f
application. find torques about a (common) axis
Note: can have
Fnet .eq. 0
• May need to apply Second Law twice to each part
G
G
but τnet .ne. 0
¾ Translation:
Fnet = Fi = ma
∑
¾ Rotation:
G
G
G
τnet = ∑ τi = Iα
• Make sure there are enough
g (N) equations;
q
there may
y be constraint
equations (extra conditions connecting unknowns)
• Simplify and solve the set of (simultaneous) equations.
• Interpret the final formulas. Do they make intuitive sense? Refer back
to the sketches and original problem
• Calculate numerical results, and sanity check anwers (e.g., right order of
magnitude?)
Example: Torque and Angular Acceleration of a Wheel
Analysis approach:
ƒ
Break into two sub-systems
‰ wheel is accelerated angularly by tension T
‰ block is accelerated linearly by weight mg, with tension
opposing
ƒ
Free body diagrams shown
ƒ
The wheel is rotating and so we apply
Στ = Ια
r
a
‰ The tension supplies the torque via tangential force
ƒ
ƒ
The mass is moving in a straight line, so apply
Newton’s
Newton
s Second Law
ΣFy = may = mg -T
How to connect the two problems above?
mg
‰ Need a constraint linking linear acceleration to α
a = αr
4
Application of Method using the Second Law
• Cord wrapped around disk, hanging weight
• Cord does not slip or stretch Æ constraint
• Disk’s rotational inertia slows accelerations
• Let m = 1.2 kg, M = 2.5 kg, r =0.2 m
•For
Findmass
acceleration
of mass m, find α for disk
m:
T
y
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
∑ τ0 =
α=
T
Mg
+ Tr = Iα
Tr m(g − a)r
= 1
I
Mr 2
I=
1 2
Mr
2
Unknowns: a, α
So far: 2 Equations, 3 unknowns ÆNeed a constraint:
Substitute and solve:
α=
mg
2
2mgr 2mαr 2
Mr 2
Mr 2
m
2mg
α(1 + 2 ) =
M
Mr
from “no
slipping”
assumption
a = + αr
α=
mg
(= 24 rad/s 2 )
r(m + M/2)
a=
mg
(= 4.8 m/s 2 )
(m + M/2)
α proportional
to g
Example: Heavy pulley Atwood’s machine
application of Newton’s 2nd Law
Given the numerical values:
• m1= 5.0 kg,
• m2= 5.5 kg
• r = 0.2 m,
• M = 6.0
6 0 kg
• I = 0.20 kg-m2
a)
b)
c)
d)
Find the accelerations of the hanging
weights. Are they up or down?
Find the angular acceleration of the pulley.
Is it CW or CCW?
Find the tensions in each cord.
H
How
llong does
d
it ttake
k th
the 5
5.5
5 kg
k weight
i ht to
t
fall 0.6 m from rest?
r
O
M, I
m2
m1
Strategy:
Apply Newton’s Second Law (linear and rotational form)
to all three bodies, connected by constraints. Make sure there
are as many independent equations as there are unknowns
5
Heavy pulley Atwood’s machine: solution
Apply Second Law:
∑ F i = ma
= ∑ τi = I α
(I) Fnet =
(each component)
(II) τ net
(pulley)
N
If I = 0 problem simplifies
O
For m1 use I:
m1g
m2
• positive α Æ negative a1 (down)
For m2 use I again:
m 2 a 2 = T2 - m 2 g
m1
• choose y positive up for
both masses
• CCW rotation Æ positive α
T2
T2 = m 2 ( a 2 + g)
a2
m2g
• positive α Æ positive a2 (up)
• Free fall if T1 or T2 = 0
Constraints: cord cannot stretch or slip
a 2 = − a1
a1 = − αr
Note: if I = 0, then T1 = T2.
Result using 1 & 2 becomes:
only 2 of these
are independent
a2 =
Atwood’s solution, continued
For the Pulley use II:
τ net = + T1r − T2ri
3. α =
y
Mg
y
a1
1. T1 = m 1 ( a 1 + g)
a 2 = + αr
M, I
T1
m 1 a 1 = T1 - m 1g
2.
α
r
m1 - m 2
g
m1 + m 2
N
= I α
( T1 − T2 ) r
I
O
1
Check ability to solve:
T1 ≠ m 1g
T2 ≠ m 2 g
M, I
T
T
• p
positive α Æ T1 > T2
NOTE:
α
r
M
Mg
2
• 5 unknowns: a1, a2, T1, T2, α
• Have 5 independent equations: 1,2,3 and 2 constraints
OK!
Solution: Begin eliminating unknowns (tension equations)
I α = T1r − T2ri = (m 1g + m 1a 1 - m 2 a 2 - m 2 g)r = (m 1 - m 2 )gr + (m 1 + m 2 )a 1r
I α - (m 1 + m 2 )a 1r = (m 1 - m 2 )gr
I α + (m 1 + m 2 ) α r 2 = (m 1 - m 2 )gr
Result:
l
interpretation
is simple
Limiting case:
Let I Æ 0:
massless
pulley
α=
α→
(m 1 − m 2 ) g r
I + (m 1 + m 2 ) r
(m 1 − m 2 ) g
(m 1 + m 2 ) r
a 2 = − a1
a 1 = − αr
Net torque of hanging weights
2
Total moment of Inertia
a1 = − αr = −
m1 - m 2
g
m1 + m 2
Negative if m1 > m2:
6
Atwood’s solution, numerical evaluation
α=
( 5 . 0 − 5.5 ) ( 9 .8) (0.2)
0.20 + ( 5 . 0 − 5.5 ) 0.2
T1 = 5.0 (9.8 + 0.32)
T2 = 5.5 (9.8 − 0.32)
2
(b)
α = - 1.58 rad/s 2 ( CW )
(a)
a 1 = - α r = + 0.32 m/s 2 (up)
a 2 = - a 1 = − 0.32
0 32 m/s 2 (down)
(c)
T1 = 506 N.
T2 = 521 N.
(a 1 increases tension)
(a 2 decreases tension)
How long does m2 take to fall 0.6 m from rest?
d=
1
a 2t 2
2
(d)
t=
2d
= 1.94 s
a2
7