Physics 215 – Fall 2014 Lecture 10-2 1

Welcome back to Physics 215

Today’s agenda:

• Extended objects

• Torque

• Moment of inertia

Physics 215 – Fall 2014 Lecture 10-2 2

Current homework assignment

• HW8:– Knight Textbook Ch.12: 8, 54, 58, 60, 62, 64– Due Monday, Nov. 5th in recitation

Physics 215 – Fall 2014 Lecture 10-2 3

Motion of center of mass of a system:

The center of mass of a system of point objects moves in the same way as a single object with the same total mass would move under the influence of the same net (external) force.

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Throwing an extended object

• Demo: odd-shaped object – 1 point has simple motion = projectile motion - center of mass

• Total external force Fext

aCM = Fext /M

• Translational motion of system looks like all mass is concentrated at CM

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Equilibrium of extended object

• Clearly net force must be zero

• Also, if want object to behave as point at center of mass ALL forces acting on object must pass through CM

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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.

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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.

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Preliminary 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.

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Computing torque

d

F= |F|d = |F||r|sin= (|F| sinr|

component of force at 900 toposition vectortimes distance

r

O

Physics 215 – Fall 2014 Lecture 10-2 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 is the angle between the vectors r and F.

Physics 215 – Fall 2014 Lecture 10-2 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 AB is perpendicular to the plane of vectors A and B and given by the right-hand rule.

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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 (-) about pivot point

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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.

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1. Clockwise.

2. Counter-clockwise.

3. Not at all.

4. Not sure what will happen.

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?

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Examples of stable and unstable rotational equilibrium

• Wire walker – no net torque when figure vertical.

• Small deviations lead to a net restoring torque stable

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Restoring torque

dR

dL

Consider displacinganticlockwise

R increasesL decreases

net torque causesclockwise rotation!

e.g., wire walker

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Rotations about fixed axis• Every particle in body undergoes

circular motion (not necessarily constant speed) with same time period

• v = (2r)/T = r. Quantity is called angular velocity

• Similarly can define angular acceleration t

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Rotational Motion

pivot

ri|vi| = ri at 90º

to ri mi

* Particle i:

Fi

* Newton’s 2nd law:

mivi/t = FiT component

at 90º to ri

miri2/t = Fi

T

ri = i* Finally, sum over all masses:

* Substitute for vi and multiply by ri :

/t) miri2 = i =

net

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Discussion

/t) miri2

= net angular acceleration Moment of inertia, I

I= net

compare this with Newton’s 2nd law

M a = F

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Moment of Inertia* I must be defined with respect to a particular axis

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Moment of Inertia of Continuous Body

r m

m a 0

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Tabulated Results for Moments of Inertia of

some rigid, uniform objects

(from p.299 of University Physics, Young & Freedman)

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Parallel-Axis Theorem

CM

D

D

*Smallest I will always be along axis passing through CM

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Practical Comments on Calculation of Moment of Inertia for Complex Object

1. To find for a complex object, split it into simple geometrical shapes that can be found in Table 9.2

2. Use Table 9.2 to get CMfor each part about the axis parallel to the axis of rotation and going through the center-of-mass

3. If needed use parallel-axis theorem to get for each part about the axis of rotation

4. Add up moments of inertia of all parts

Physics 215 – Fall 2014 Lecture 10-2 27

Beam resting on pivot

M = ? mMb = 2m

NCM of beam

Vertical equilibrium?

Rotational equilibrium?

r rx

rm

F =

=

M =N =

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Suppose M replaced by M/2 ?

• vertical equilibrium?

• rotational dynamics?

• net torque?

• which way rotates?

• initial angular acceleration?

F =

=

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Moment of Inertia?

I = miri2

* depends on pivot position!

I =

* Hence

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Rotational Kinetic Energy

K = i(mivi2 = (1/2)imiri

2

• Hence

K = (1/2)I

• This is the energy that a rigid body possesses by virtue of rotation

Physics 215 – Fall 2014 Lecture 10-2 31

Spinning a cylinderCable wrapped aroundcylinder. Pull off withconstant force F. Supposeunwind a distance d of cable

F

• What is final angular speed of cylinder?

2R

• Use work-KE theorem

W = Fd = Kf = (1/2)I 2

• Mom. of inertia of cyl.? -- from table: (1/2)mR2

– from table: (1/2)mR2

= [2Fd/(mR2/2)]1/2 = [4Fd/(mR2)]1/2

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Reading assignment

• Rotations

• Chapter 12 in textbook