Foundation course - PHYSICS
Lecture 6-2: Momentum; rotational motion
Dynamics of solid bodies
● Center of mass
● Linear momentum
● Impulse
● Rotational motion
● Torque
● Angular momentum
Naďa Špačková spackova@physics.muni.cz
Center of mass
The center of mass of a
system of particles is the point
that moves as though:
● all of the system’s mass
were concentrated there
● all external forces were
applied there
System of two particles: xcom =
m1 x1 + m2 x2
m1 + m2
The center of
mass position:
Center of mass
The center of mass (COM):
● all of the system’s mass were concentrated there
● all external forces were applied there
xcom =
m1 x1 + m2 x2
m1 + m2
The center of mass position:
Composite particle
Linear momentum
Linear momentum of a particle is a vector quantity that is defined as:
p = m v
Newton’s second law expressed in terms of momentum:
Fnet =
d p
d t
Fnet = m
d v
d t
= m a
System of two and more particles:
P = p1+ p2 = m1 v1 + m2 v2
P = M vcom
Fnet = M acom
M is a total mass of the system
Collision and impulse
Collision:
The external force acting
on a body is brief, has
large magnitude, and
suddenly changes the
body’s momentum
Impulse:
The impulse on an object is
the product of the average
force on an object and the
time interval over which it
acts.
Impulse-momentum theorem
FΔt = mΔ v = m vf −m vi = pf −pi = Δ p
The impulse on an object is equal to the change in its momentum.
Impulse-momentum theorem: J = Favg Δt = Δ p
● a large impulse causes a large change in momentum
● the large impulse could result either from a large F acting over a short
Δt or from a smaller F acting over a longer Δt
Example: Air bags in cars reduce injuries by making the force on an passenger
less, by increasing the time interval of force acting and by spreading the force over
a larger area of the person’s body.
Newton’s second law of motion: F = m a = m(Δ v
Δt )
Conservation of momentum
Collision of two balls (closed and isolated system):
FDC =−FCD (FΔt)DC =−(F Δt)CD
pCf − pCi =−( pDf −pDi) pCf + pDf = pCi+ pDi
Law of conservation of momentum:
Momentum of any closed, isolated system
does not change.
vC
vD
A system with conserved mass = closed system
A system with the zero net external force (only internal
forces are included) = isolated system
P = constant
Checkpoint question:
Collisions: momentum and kinetic energy
Closed and isolated system: momentum of the system is constant
Collision is elastic: kinetic energy of the system is conserved
Collision is inelastic: kinetic energy of the system is not conserved
Inelastic collisions in one dimension
m1 v1i + m2 v2i = m1 v1f + m2 v2f
p1i + p2i = p1f + p2 f
m1 v1i = (m1 + m2)V
Collisions: momentum and kinetic energy
Elastic collisions in one dimension
p1i + p2i = p1f + p2 f
Ek1i + Ek 2i = Ek1f + Ek 2f
1
2
m1 v1i
2
=
1
2
m1 v1f
2
+
1
2
m2 v2f
2
pA
pA’
pB’
pA
pA’
pB’
Collisions in two dimensions
com
m1 v1i = m1 v1 f + m2 v2f
Circular motion
r1
r2
r1
v1
v2
v1
v2 Δv
a
a
¯a =
Δ v
Δt
Average acceleration:
Centripetal acceleration: ac=
v2
r
¯v =
Δ r
Δt
Δr
Average velocity:
Period of revolution T:
● time needed for the object to make one complete revolution
● during this time the object travels a distance equal to the circumference
of the circle (2πr)
v =
2πr
T
ac =
(2πr/T )
2
r
=
4 π
2
r
T
2
Centripetal force
Centripetal force
● because the acceleration of an object is always in the direction of the net
force acting on it, the net force must be toward the center of the circle
Examples:
● Earth circling the Sun – Fc
is Sun’s
gravitational force
● Hammer thrower swings the hammer – Fc
is
the tension in the chain attached to the ball
Fc= mac
Newton’s second law for circular motion
r
Fc
v
v
a
Fc=
m v2
R
Angular displacement
Fraction of one revolution can be measured:
● in degrees (one complete revolution is 360°)
● in radians (one complete revolution is 2π)
Radian is related to the ratio of the circumference
of a circle to its radius.
Angular displacement θ:
● the change in the angle if an object
rotates
Measuring distance:
● one complete revolution …x = 2πr
● generally for an angle θ … x = θ r
Radians are dimensionless.
Clockwise rotation is negative, counterclockwise
rotation is positive.
θ =
x
r
Angular velocity
● is defined as and angular displacement divided by the time taken
to make the angular displacement
Average angular velocity ω:
ω = Δθ
Δt
Instantaneous angular velocity equals the slope
of a graph of angular position versus time.
Linear velocity of a point at a distance
r from the axis of rotation:
¯v =
Δ x
Δt
=
r Δθ
Δt
= r ω
Earth’s angular velocity
Direction of angular velocity:
the right-hand rule
Angular acceleration
Angular acceleration is the change in angular velocity divided by the
time required to make the change
Average angular acceleration α:
α = Δ ω
Δt
Linear and angular measurements
Quantity Linear Angular Relationship
displacement x (m) θ (rad) x = r θ
velocity v (m/s1
) ω (rad/s1
) v = r ω
acceleration a (m/s2
) α (rad/s2
) a = r α
Angular frequency:
number of complete revolutions made
by an object in 1 s
f ≡ ω
2 π
Acceleration
Linear speed:
Tangential acceleration at
:
at = αr
Radial acceleration ar
:
ar =−
v
2
r
=−ω2
r
v = ωr
Kinetic energy of rotation
Moment of inertia I:
Rotational kinetic energy:
Ek =
1
2
m1 v1
2
+
1
2
m2 v2
2
+ ... =∑ 1
2
mi vi
2
Ek =
1
2
m v
2
This equation is valid only for a particle
Rigid body is a collection of particles with different speeds:
Ek = ∑ 1
2
mi vi
2
= ∑ 1
2
mi(ωri)
2
=
1
2
(∑mi r1
2
)ω
2
I = ∑mi r1
2
Ek =
1
2
I ω
2
The moment of inertia characterizes
the resistance to rotation
I=mr
2
Moment of inertia of a point mass:
The moment of inertia
Parallel-axis theorem: I = Icom + M h
2 h...distance between parallel axis and
axis through the center of mass
Checkpoint question:
Checkpoint question:
Torque
How to open a door most easily – how to get the most effect from the
least force
Application of the force farthest
from the hinges is most effective.
Application of the force at an angle
perpendicular to the door is most
effective.
Torque
Torque τ:
● quantity characterizing the ability of the force F to rotate the body
● it depends on the magnitude of Ft
and how far from O the force is
applied
τ = r Fsinϕ
Torque is:
● is a vector
● is a vector product of r and F
● measured in units N.m
τ = r×F
τ = r Fsinϕ
Two equivalent ways of computing the torque:
τ =(r)(F sinϕ)= r Ft τ =(r sinϕ)(F)= r⊥ F
Torque
Checkpoint question:
Torque
Direction of torque:
● clockwise rotation = torque is negative
● counterclockwise rotation = torque is positive
τ = r×F
When several torques act on a body, the net torque is the sum of the
individual torques.
Torque is a vector and a vector product of r and F:
Newton’s second law for rotation:
Fnet = ma τnet = I α
τ = Ft r = mat r = m(α r)r =(mr2
)α = I α
Finding net torque
Short lever arm requires to exert
big force
Long lever arm requires to exert
less force
r1
r2
F1
F2
τN
= τ1
+ τ2
r1
= r2
and F1
= F2
τN
= F1
r1
– F2
r2
= 0
System is in
equilibrium and
does not rotate
Angular momentum
τ = I α = I Δω
Δt
τ Δt = I Δω = I ωf −I ωi
Angular momentum L: L = I ω
The angular momentum is defined as a product of the object’s moment
of inertia and the object’s angular velocity.
units: 1 kg.m2
.s-1
Corresponding variables:
Translational motion Rotational motion
Position x Angular position θ
Velocity v Angular velocity ω
Acceleration a Angular acceleration α
Mass m Moment of inertia I
Force F Torque τ
Linear momentum p Angular momentum L
Conservation of angular momentum
An isolated system’s initial angular momentum is equal to its final angular
momentum.
Li = Lf L = I ω … is constant
angular velocity: ω1
< ω2
Increased angular velocity is accompanied
by a decreased moment of inertia.
I∼r
2
Moment of inertia can be decreased by
decreasing the radius of rotation
The direction of rotation of a spinning
object can be changed only by applying
a torque.
L = constant