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Example
20m 40m
A B
€
rτ net, A =
r τ T, A +
r τ B, A
0 = (±)FT rT + (±)FB rB
0 = −FT rT + FB rB
FB =FT rT
rB
=8000(20)
(60)= 2667 N
A truck crosses a massless bridge supported by two piers. What force much each pier exert when the truck is at the indicated position?
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rF A
€
rF B
€
rF T = 8000 N
€
rτ net, B =
r τ T, B +
r τ A, B
0 = (±)FT rT + (±)FA rA
0 = +FT rT − FA rA
FA =FT rT
rA
=8000(40)
(60)= 5333 N
€
2667 + 5333 = 8000 N
Center of MassUntil now we’ve been treating objects as though they were single point masses and ignoring any rotation or other motion that the object may have.
We can do this because there is a point on (or near) the object that does describe the kinds of motions that we’ve been investigating.
This point is the center of mass.
For Uniform Objects
The center of mass is at the geometric center (the balance point).
Experimentally – this is easy to find.
Simple CaseThe center of mass of an object is a “weighted” average of the position of all the pieces that make up that object.
Each piece’s position is weighted by what fraction of the total mass that piece contains.
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xCM =m1
mtotal
x1 +m2
mtotal
x2 +K =1
mtotal
mix ii
∑
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+x
1 m 2 m
3 kg 6 kg 1 kg
€
xCM =m1x1 + m2x2 + m3x3
m1 + m2 + m3
=3(−1) + 6(0) +1(2)
3 + 6 +1= −
1
10
Simple CaseThe center of mass of an object is a “weighted” average of the position of all the pieces that make up that object.
Each piece’s position is weighted by what fraction of the total mass that piece contains.
€
xCM =m1
mtotal
x1 +m2
mtotal
x2 +K =1
mtotal
mix ii
∑
€
+x
1 m 2 m
3 kg 6 kg 1 kg
€
xCM =m1x1 + m2x2 + m3x3
m1 + m2 + m3
=3(0) + 6(1) +1(3)
3+ 6 +1=
9
10
Oddly Shaped ObjectsHere the center of mass may not even lie “on” the object and we may need to resort to a calculation in 2D
Mass X coord Y coord
2 kg 4m 4m
3 kg 0m 0m
4 kg 2m -3m
1 kg 4.5m -1m
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yCM =m1y1 + m2y2 + m3y3 + m4 y4
m1 + m2 + m3 + m4
=2(4) + 3(0) + 4(−3) +1(−1)
2 + 3+ 4 +1= −0.5
€
xCM =m1x1 + m2x2 + m3x3 + m4 x4
m1 + m2 + m3 + m4
=2(4) + 3(0) + 4(2) +1(4.5)
2 + 3 + 4 +1=1.25
Stability
Stable Not Stable
An object is said to be stable when its center of mass is directly over its base of support.
In such cases, any slight movement of the CM away from the equilibrium position produces a torque that brings the object back to equilibrium.
Stability and TorqueAn 80 kg pirate is being forced to walk the plank. The plank is a wooden beam of mass 240 kg and is 20 m long. 8 m of the beam sticks out beyond the edge of the wall. How far can the pirate walk out onto the beam before it falls?
20m
8m
The plank becomes unstable and begins to rotate when the CM is at the edge of the wall.
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xCM =mb xb + mp x p
mb + mp
x p =xCM( ) mb + mp( ) − mb xb
mp
=12( ) 240 + 80( ) − 240( ) 10( )
80 =18 m
€
+x
Stability and TorqueAn 80 kg pirate is being forced to walk the plank. The plank is a wooden beam of mass 240 kg and is 20 m long. 8 m of the beam sticks out beyond the edge of the wall. How far can the pirate walk out onto the beam before it falls?
20m
8m
What is the torque on the board if the pirate stands at x = 18.5 m? Take the corner of the wall to be the origin.
€
+x
€
rF g,b
€
rF p,b
€
τnet = τ p +τ g
= rpFp, perp + rgFg, perp
= −rp mpg( ) + rCM,b mbg( )
= −6.5 80*9.8( ) + 2 240*9.8( )
= −392 Nm
Newton’s LawsNewton’s 2nd Law – If the net torque on an object about a point is not zero, then the
net torque produces an angular acceleration about that point.
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rF
€
rr
€
τ =rF
= r ma( )
= r m rα( )( )
= mr2α
The quantity mr2 is the rotational equivalent of mass, and is called moment of inertia.
For a rigid body, there can be many masses (m’s) and many distances (r’s).
O
€
I ≡ miri2
i
∑
€
=mr2α = Iα
Example Rotational Inertia
3m 6m
4 kg 2 kg
Rotational axis
€
I = miri2
i
∑
= m1r12 + m2r2
2
= 4 32( ) + 2 62
( ) =108 kg m2
3m6m
4 kg 2 kg
Rotational axis
Example Rotational Inertia
€
I = miri2
i
∑
= m1r12 + m2r2
2
= 4 62( ) + 2 32
( ) =162 kg m2
€
I = miri2
i
∑
= m1r12 + m2r2
2
= 4 32( ) + 2 62
( ) =108 kg m2
3m
6m
4 kg 2 kg
Rotational axis
Example Rotational Inertia
€
I = miri2
i
∑
= m1r12 + m2r2
2
= 4 62( ) + 2 32
( ) =162 kg m2
€
I = miri2
i
∑
= m1r12 + m2r2
2
= 4 32( ) + 2 62
( ) =108 kg m2
4m
It is shortest distance to axis of rotation that matters!In this case, the right sphere is still only 3m away from the axis.
Determining Rotational Inertia by Experiment
33cm
15N
A 15 N force is applied to a wheel of mass 4 kg as shown. The wheel is observed to accelerate from rest to an angular speed of 30 rad/s in 3 sec. Determine the rotational inertia of the wheel.
€
τ =Iα
I =τ
α = τ Δω
Δt( )
=15 × 0.33
30−0( )3
= 0.495 kg m2
Rotational Dynamics
€
α =τ Iτ = r1F1, Perp + r2F2, Perp + r3F3, Perp + r4F4, Perp
= r1m1g + r2 0( ) − r3m3g + 0( )
= 3* 3*9.8 − 2* 4 *9.8 = 9.8 Nm
I = m1r12 + m2r2
2 + m3r32 + m4r4
2
= 3* 32 +1*22 + 4 *22 + 0 = 47 kg m2
α =9.8
47= 0.21 rad
s2
3 m 2 m
m1 = 3 kgm2 = 1 kgm3 = 4 kgm4 = 5 kg
Four spherical masses are attached by massless rods. They rotate about an axis that is perpendicular to the plane of the page and passes through the black mass. What angular acceleration does the torque due to gravity cause?
2 m
€
at = αr3
at = 0.21*2 = 0.42 ms2
What is initial tangential acceleration of m3?
€
rF g, 1
€
rF g, 4
€
rF g, 3
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rF g, 2
What about initial centripetal acceleration?
Rotational Inertia for Various Common Objects
Hoop (or Ring)Solid Disc
Solid Sphere
RR
R R
Spherical Shell
