L21-s1,11

Physics 114 – Lecture 21• §7.6 Inelastic Collisions• In inelastic collisions KE is not conserved, which is

equivalent to saying that mechanical energy is not conserved

• Collisions where the two bodies stick together after the collision are known as completely inelastic collisions

• Study Example 7.9 – the energy lost during a collision between two railroad cars

• We now study the ballistic pendulum

L21-s2,11

Physics 114 – Lecture 21• Example 7.10 – the ballistic pendulum

• Momentum conservation in

collision →

• mv = (m + M)V ´

• Mechanical energy conservation

after collision →

• ½ (m + M)V ´2 = (m + M)gh

•

• h = L – L cosθ = L (1 – cosθ)

Mm v

VM = 0

m + M V´

h2gh

m

M m V

m

M m v

L θ

L21-s3,11

Physics 114 – Lecture 21• §7.8 Center of Mass (CM)• So far we have considered the motion of point

particles, that is the motion of a body where its position can be specified precisely

• What coordinates do we use when we deal with an extended body, c.f., races in athletics, etc.?

• It turns out that, by defining a quantity known as the Center of Mass (CM), all of our equations in Newton’s Mechanics remain valid provided that we consider the motion of the center of mass of that body

L21-s4,11

Physics 114 – Lecture 21• Definition of Center of Mass

• Location of the center of mass (CM) is defined to be

• and, generalizing to two dimensions, we have for yCM,

i

ii

4321

44332211CM m

xm

.....mmmm

.....xmxmxmxmx

x0x1 x2 x3 x4 x4 …m1 m2 m3 m4 m4 …

i

ii

4321

44332211CM m

ym

...mmmm

...ymymymymy

L21-s5,11

Physics 114 – Lecture 21• Chapter 8 Rotational Motion

• §8.1 Angular Quantities

• So far we have considered point particles where no complications occurred regarding its rotation

• Once we consider the motion of an extended body we need to consider its rotational motion

• First we need to remind ourselves about angular quantities

• , where θ is expressed in radians

• or s = rθ

• Study Example 8.1 r

r

sθ

θs (along arc)

L21-s6,11

Physics 114 – Lecture 21• Rotational motion of a wheel

• The wheel rotates so that a point

on the rim moves from an initial

position, θ1, to a final position, θ2

• Δ θ = θ2 - θ1

• Define the average angular velocity,

• As before the instantaneous angular velocity is then defined as,

t

θ ω

θ1

θ2

t

θωω lim

0tinst

t

θω

L21-s7,11

Physics 114 – Lecture 21• We define, similarly, the average angular acceleration,

• and the instantaneous angular acceleration,

• Each point on the rim of the wheel

has a linear velocity,

t

ωαα lim

0tinst

t

ω

t

ωωα 12

ΔsΔθ

rrωt

θr

t

θr

t

sv

L21-s8,11

Physics 114 – Lecture 21

• Note that the expression, v = r ω is valid for the linear velocity at any radius on the wheel

• Likewise this expression holds for any rotating body, where r is the distance from the axis of rotation to the point at a distance, r, perpendicular to this axis and provided also that the object is a rigid body, that is that all points in the body rotate with the same angular velocity, ω

• Study Examples, 8.3, 8.4 and 8.5

L21-s9,11

Physics 114 – Lecture 21

• §8.2 Constant Angular Acceleration• Since the definitions of ω and α from θ and t

correspond exactly to the definitions of v and a from x and t, we have a similar set of relations between ω, α, θ and t

L21-s10,11

Physics 114 – Lecture 21

• Kinematic Formulae for Rotational Motion

• Linear Angular

• v = v0 + a t ω = ω0 + α t

• x = x0 + v0 t + ½ a t2 θ = θ0 + ω0 t + ½ α t2

• v2 = v02 + 2 a (x - x0) ω2 = ω0

2 + 2 α (θ - θ0)

• Study Example 8.6

tvx- x,2

vvv

00

tωθ -θ ,

2

ω ωω

00

L21-s11,11

Physics 114 – Lecture 21

• §8.3 Rolling Motion (Without Slipping)• v = r ω and a = r α• Study Example 8.7