Physics 106P Lecture 24 Notes

1
Elliotts Riddle
There's this hypothetical green glass door. Only
certain things exist beyond the door. There are
no politics, no laws, and no government in the
green glass world, but there is a congress. There
is no president, or kings in the green glass
world, but there is a queen. There is engineering
without physics. Pizza, spaghetti, and food
exists, but there is no pasta, corn, sushi, cake,
or sandwiches. Your mouth can't go through the
door but your teeth can. Shoes can't go through,
but feet can. There are no sports, but there is
baseball, basketball, football, and even soccer.
Just no golf.

• I see the answer
• I dont see the answer

2
Physics 211 Lecture 20Todays Agenda

• Torque due to gravity
• Rotation Recap
• Statics
• Car on a Hill
• Static Equilibrium Equations
• Examples
• Suspended beam
• Hanging lamp
• Ladder

3
Lecture 20, Act 1Rotations

• A ball and box have the same mass and are moving
  with the same velocity across a horizontal floor.
  The ball rolls without slipping and the box
  slides without friction. They encounter an
  upward slope in the floor. Which one makes it
  farther up the hill before stopping?

(a) ball (b) box (c) same

4
Lecture 20, Act 1 Solution

• The ball and box will stop when their initial
  kinetic energies have been converted to
  gravitational potential energy (mgH).

• The initial kinetic energy of the box is

• The initial kinetic energy of the ball is

bigger
v
v
w
5
Lecture 20, Act 1 Solution

• Since the ball has more initial kinetic energy,
  it will go higher!

6
Recap of Rotation so far

• About a fixed rotation axis, you can always write
  ? I? where ? is the torque, I is the moment of
  inertia, and ? is the angular acceleration.
• For discrete point particles, I ?miri2
• The parallel axis theorem lets you calculate the
  moment of inertia about an axis parallel to an
  axis through the CM if you know ICM
• If the object is accelerating, we can still use ?
  I?? provided that we are considering rotations
  about an axis through the CM.

IPARALLEL ICM MD2
7
Torque due to Gravity
hang odd objects

• As we now know where ?i ri X Fi

• Take the rotation axis to be along the z
  direction (as usual) and recall that
• ??i ?Z,i rX,i FY,i - FX,i rY,i
• xi (-mi g) - 0

y
m4
x
F4
r4
m1
z-axis
r1
F1
r2
m2
r3
m3
F2
F3
8
Torque due to Gravity...

• But this is the same expression we would get if
  we were to find the CM...

y
CM
9
Torque due to Gravity...

• ...and assume that all of the mass was located
  there!

• So for the purpose of figuring out the torque due
  to gravity, you can treat an object as though all
  of its mass were located at the center of mass.

y
M
rcm
xcm
Mg
10
New Section - Statics

• As the name implies, statics is the study of
  systems that dont move.
• Ladders, sign-posts, balanced beams, buildings,
  bridges, etc...
• Example What are all ofthe forces acting on a
  carparked on a hill?

N
f
mg
?
11
Car on Hill
Truck on hill

• Use Newtons 2nd Law FNET MACM 0
• Resolve this into x and y components

x f - mg sin ? 0 f mg sin ?
N
y N - mg cos ? 0 N mg cos ?
f
mg
?
12
Using Torque

• Now consider a plank of mass M suspended by two
  strings as shown. We want to find the tension in
  each string

T1
T2
M
x cm

• This is no longer enough tosolve the problem!
• 1 equation, 2 unknowns.
• We need more information!!

L/2
L/4
Mg
13
Using Torque...

• We do have more information
• We know the plank is not rotating!
• ?NET I? 0

T1
T2
M
x cm
L/2
L/4

• The sum of all torques is zero!
• This is true about any axiswe choose!

Mg
14
Using Torque...

• Choose the rotation axis to be along the z
  direction (out of the page) through the CM

T1
T2
M
x cm
L/2
L/4
Mg
Gravity exerts notorque about the CM
15
Using Torque...
Suspended beam

• Since the sum of all torques must be 0

T1
T2
M
x cm
L/2
L/4
Mg
16
Approach to Statics
Torque equilibrium

• In general, we can use the two equations
• to solve any statics problem.
• When choosing axes about which to calculate
  torque, we can be clever and make the problem
  easy....

17
Lecture 20, Act 2Statics

• A 1 kg ball is hung at the end of a rod 1 m long.
  The system balances at a point on the rod 0.25 m
  from the end holding the mass.
• What is the mass of the rod?

(a) 0.5 kg (b) 1 kg (c) 2 kg

1 m
1 kg
18
Lecture 20, Act 2Solution A

• The total torque about the pivot must be zero.

1 kg
19
Lecture 20, Act 2Solution B

• Since the system is not rotating, the
  x-coordinate of the CM of the system must be the
  same as the pivot.

1 kg
x
20
Example Hanging Lamp

• A lamp of mass M hangs from the end of plank of
  mass m and length L. One end of the plank is held
  to a wall by a hinge, and the other end is
  supported by a massless string that makes an
  angle ? with the plank. (The hinge supplies a
  force to hold the end of the plank in place.)
• What is the tension in the string?
• What are the forces supplied by thehinge on the
  plank?

21
Hanging Lamp...

• First use the fact that in both x
  and y directions

x T cos ? Fx 0 y T sin ? Fy - Mg - mg
0
T
Fy
??
m
Fx
L/2
L/2
M
mg
Mg
22
Hanging Lamp...

• So we have three equations and three unknowns
• T cos ? Fx 0
• T sin ? Fy - Mg - mg 0

which we can solve to find
T
Fy
??
m
Fx
L/2
L/2
M
mg
Mg
23
Lecture 20, Act 3Statics

• A box is placed on a ramp in the configurations
  shown below. Friction prevents it from sliding.
  The center of mass of the box is indicated by a
  blue dot in each case.
• In which cases does the box tip over?

(a) all (b) 2 3 (c) 3 only

3
1
2
24
Lecture 20, Act 3 Solution

• We have seen that the torque due to gravity acts
  as though all the mass of an object is
  concentrated at the center of mass.

• If the box can rotate in such a way that the
  center of mass islowered, it will!

3
1
2
25
Lecture 20, Act 3 Solution

• We have seen that the torque due to gravity acts
  as though all the mass of an object is
  concentrated at the center of mass.

• Consider the bottom right corner of the box to be
  a pivot point.

• If the box can rotate in such a way that the
  center of mass islowered, it will!

3
1
2
26
Example Ladder against smooth wall

• Bill (mass M) is climbing a ladder (length L,
  mass m) that leans against a smooth wall (no
  friction between wall and ladder). A frictional
  force F between the ladder and the floor keeps it
  from slipping. The angle between the ladder and
  the wall is ?.
• What is the magnitude of F as a function of
  Bills distance up the ladder?

?
L
m
Bill
F
27
Example Ladder against smooth wall...

• Consider all of the forces acting. In addition to
  gravity and friction, there will be normal forces
  Nf and Nw by the floor and wall respectively on
  the ladder.
• Again use the fact that FNET 0 in
  both x and y directions
• x Nw F
• y Nf Mg mg

Nw
L/2
?
m
mg
d
Mg
F
Nf
28
Example Ladder against smooth wall...

• Since we are not interested in Nw, calculate
  torques about an axis through the top end of the
  ladder, in the z direction.

axis
Nw
L/2
?
m

• Substituting in Nf Mg mg andsolving for F

mg
d
Mg
F
Nf
29
Example Ladder against smooth wall...
Ladder vs. wall

• We have just calculated that
• For a given coefficient of static friction
  ?s,the maximum force of friction F that can
  beprovided is ?sNf ?s g(M m).
• The ladder will slip if F exceedsthis value.
• Morals
• Brace the bottom of ladders!
• Dont make ? too big!

?
m
d
F
30
Recap of todays lecture

• Torque due to gravity Rotation Recap
• Statics
• Car on a Hill
• Static Equilibrium Equations
• Examples