Physics 201: Lecture 19

1
Physics 201 Lecture 19

• Dynamics
• Reprise
• Statics
• Car parked on a hill
• Static Equilibrium Equations
• Examples
• Suspended beam
• Hanging lamp
• Ladder

2
Summary of Dynamics

• Dynamics (cause effect)
• Force Linear acceleration
• Change in linear momentum, Fnet dp/dt
• Torque angular acceleration
• Change in angular momentum, tdL/dt
• Conservation of momentum
• When net external force zero
• Conservation of angular momentum
• When net external torque zero
• Conservation of Energy
• Mechanical energy - kinetic potential
• Non-conservative forces e.g. friction
• Energy lost to the environment, e.g., heat

3
Dynamical Reprise Linear

• The change of motion of an object is described by
  Newton as Fma where F is the force, m is the
  mass, and a is the acceleration.
• For discrete point particles, all forces act on
  the center of mass
• The center of mass is a calculable property of
  the object.

If F0, then there is no change of motion -- if
at rest, the object will remain at rest.
4
Dynamical Reprise Rotation

• 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.

5
Question

• 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?

(1) ball (2) box (3) same

6
Solution

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

v
v
w
7
Solution

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

8
Torque due to Gravity

• As we now know where

• Take the rotation axis to be along the z
  direction (as usual) and recall that

y
m4
x
F4
r4
m1
z-axis
r1
F1
r2
m2
r3
m3
F2
F3
9
Torque due to Gravity...

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

y
CM
10
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
11
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
  car(mass m) parked on a hill?

N
f
mg
?
12
Car Parked on a 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
?
13
Scaffold

• 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!!

14
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
15
Using Torque...

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

• The torque due to the string on the right about
  this axis is

T1
T2
M
x cm
L/2
L/4

• The torque due to the string on the left about
  this axis is

Mg
Gravity exerts notorque about the CM
16
Using Torque...

• Since the sum of all torques must be 0

T1
T2
M
x cm
We already found that T1 T2 Mg
L/2
L/4
Mg
17
Approach to Statics

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

18
Question

• A 1 kg ball is hung at the end of a uniform 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?

(1) 0.5 kg (2) 1 kg (3) 2 kg

1 m
1 kg
19
Solution

• The total torque about the pivot must be zero.

1 kg
20
Another Solution

• 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
21
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?

22
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
23
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
24
Problem

• 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
  dot in each case.
• In which cases does the box tip over?

(1) all (2) 2 3 (3) 3 only

3
1
2
25
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
26
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