Welcome back to Physics 211

1
Welcome back to Physics 211

• Todays agenda
• Equilibrium of extended bodies
• Torque
• Rotational motion

2
Reminder

• Deadline for MPHW 5 extended to Wednesday,
  November 9 at noon
• MPHW 5 will be graded, despite message on MP
  website
• Lowest MPHW score will be dropped
• Professor Catterall will give lecture on Thursday
• Exam 3 in class Thursday, Nov 17

elastic energy, linear momentum, collisions,
center of mass, equilibrium of rigid bodies,
torque, rotational dynamics, angular momentum
3
A few properties of the center of mass of an
extended object

• The weight of an entire object can be thought of
  as being exerted at a single point, the center of
  mass.
• One can locate the center of mass of any object
  by suspending it from two different points and
  drawing vertical lines through the support
  points.
• Equilibrium can be ensured if all forces pass
  through CM
• An object at rest on a table does not tip over if
  the center of mass is above the area where it is
  supported.

4
A 1-kg mass is fastened to a meter stick near one
end. A person balances the system by placing a
finger directly below point P which is just to
the left of the mass. Is the center of mass of
the system located

• 1. to the left of point P,
• 2. at point P, or
• 3. to the right of point P?
• 4. Unable to decide.

5
A meterstick is pivoted at its center of mass.
It is initially balanced. A mass of 200 g is
then hung 20 cm to the right of the pivot point.
Is it possible to balance the meter-stick again
by hanging a 100-g mass from it?

• 1. Yes, the 100-g mass should be 20 cm to the
  left of the pivot point.
• 2. Yes, but the lighter mass has to be farther
  from the pivot point (and to the left of it).
• 3. Yes, but the lighter mass has to be closer to
  the pivot point (and to the left of it).
• 4. No, because the mass has to be the same on
  both sides.

6
Meter stick and mass demo

• For equilibrium need to arrange for center of
  mass to lie at pivot.
• take pivot as origin of coordinate system

7
Restatement of equilibrium conditions

• m1r1m2r2 0 ? W1r1 W2r2 0
• Thus, S force (W) x displacement (r) 0
• quantity force x displacement
• called torque (preliminary definition)
• Thus, equilibrium requires net torque to be zero

8
Tentative definition of torque
The torque on an object with respect to a given
pivot point and due to a given force is defined
as the product of the force exerted on the object
and the moment arm. The moment arm is the
perpendicular distance from the pivot point to
the line of action of the force.
9
Computing torque

• tFd
• component of
• force at 900 to
• position vector
• times distance

F
q
r
d
O
10
Definition of torque
where r is the vector from the reference point
(generally either the pivot point or the center
of mass) to the point of application of the force
F.
where q is the angle between the vectors r and F.
11
Vector (or cross) product of vectors
The vector product is a way to combine two
vectors to obtain a third vector that has some
similarities with multiplying numbers. It is
indicated by a cross (?) between the two
vectors. The magnitude of the vector cross
product is given by The direction of the
vector A?B is perpendicular to the plane of
vectors A and B and given by the right-hand rule.
12
Scalar (or dot) product of vectors
The scalar product is a way to combine two
vectors to obtain a number (or scalar) that has
some similarities with multiplying numbers (i.e.,
a product). It is indicated by a dot () between
the two vectors.
13
Interpretation of torque

• Measures tendency of any force to cause rotation
• Torque is defined with respect to some origin
  must talk about torque of force about point X,
  etc.
• Torques can cause clockwise () or anticlockwise
  rotation (-)

14
Demo fighting torque!

• hold bar add weights at different distances
• effort increases with distance and magnitude of
  weight force

15
Conditions for equilibrium of an extended object

• For an extended object that remains at rest and
  does not rotate

• The net force on the object has to be zero.
• The net torque on the object has to be zero.

16
Extended objectsneed extended free-body diagrams

• Point free-body diagrams allow finding net force
  since points of application do not matter.
• Extended free-body diagrams show point of
  application for each force and allow finding net
  torque.

17
What if t not zero ?

• If the torque about some pivot point is not zero,
  the object will rotate about the pivot.
• Rotation is consistent with direction of force

18
A T-shaped board is supported such that its
center of mass is to the right of and below the
pivot point.
Which way will it rotate once the support is
removed?

• 1. Clockwise.
• 2. Counter-clockwise.
• 3. Not at all.
• 4. Not sure what will happen.

19
Extended objectsneed extended free-body diagrams

• Point free-body diagrams allow finding net force
  since points of application do not matter.
• Extended free-body diagrams show point of
  application for each force and allow finding net
  torque.

20
(No Transcript)
21
(No Transcript)
22
(No Transcript)
23
Examples of stable and unstable rotational
equilibrium

• Wire walker no net torque when figure vertical.
• Small deviations lead to a net restoring torque ?
  stable

24
Restoring torque

• Consider displacing
• anticlockwise
• ?tR increases
• ?tL decreases
• net torque causes
• clockwise rotation!

e.g. wire walker
dR
dL
25
Rotations about fixed axis

• Every particle in body undergoes circular motion
  (not necessarily constant speed) with same time
  period
• v (2pr)/T w r. Quantity w is called angular
  velocity
• Similarly can define angular acceleration
  a??Dw/Dt

26
Rotational Motion
w
Particle i
ri
vi ri w at 90o to ri
Fi
pivot
mi
Newtons 2nd law
miDvi/DtFiT ? component at 900 to ri
Substitute for vi and multiply by ri
Finally, sum over all masses
27
Discussion
Dw/Dt (S miri2) tnet
a - angular acceleration
Moment of inertia, I
I a tnet compare this with Newtons 2nd law Ma
F
28
Moment of Inertia
I must be defined with respect to a particular
axis
29
Demo

• Spinning a weighted bar moments of inertia

30
Moment of Inertia of Continuous Body
Dm a 0
31
Tabulated Results for Moments of Inertia of some
rigid, uniform objects
(from p.342 of University Physics, Young
Freedman)
32
Parallel-Axis Theorem
CM
Smallest I will always be along axis passing
through CM
33
Practical Comments on Calculation of Moment of
Inertia for Complex Object

• To find I for a complex object, split it into
  simple geometrical shapes that can be found in
  Table 9.2
• Use Table 9.2 to get ICM for each part about the
  axis parallel to the axis of rotation and going
  through the center-of-mass
• If needed use parallel-axis theorem to get I for
  each part about the axis of rotation
• Add up moments of inertia of all parts