Motion IV
Rotary Motion
Moment/Torque
• The moment or torque is the tendency of a
force to rotate a body about a specified axis or
point.
• Moment is denoted by the vector M
• Torque is denoted by the vector τ
• Mo = r x F
• τo = r x F
F
r
o
• The vector r is any vector from the point o to
any point on the line of action of the force F.
• The cross product is a vector that is
perpendicular to both of the vectors making
up the cross product, i.e.
– The moment or torque vector is perpendicular to
both r and F
Newton’s Second Law
• For translation, F = ma
• The net external force F acting on a body of
mass m imparts an acceleration a.
• For rotation, τ = Iα
• The net external torque or moment τ acting
on a body whose mass moment of inertia is I
imparts an angular acceleration α.
Definitions
• The angular position is given by the vector ϴ
– It must be expressed in radians (not degrees).
• The angular velocity is given by the vector ω
– It must be expressed in radians/sec
• The angular acceleration is expressed by the
vector α
– It must be expressed in radians/sec2
• ω = dϴ/dt
• The angular velocity is the derivative with
respect to time of the angular position.
Radians/sec
• α = dω/dt
• The angular acceleration is the derivative with
respect to time of the angular velocity.
Radians/sec2
Equations of Motion
• There are basically three equations of rotary
motion similar to the three equations of
translational motion.
• ϴ = ϴo + ωot + (1/2)αt2
• ω = ωo + αt
• ω = ωo2 + 2α(ϴ - ϴo)
Tangential Velocity
• The tangential velocity is equal to the product
of the radius times the angular velocity, i.e.
• vt = rω
r
vt
Class Activity
• In ancient times, vinyl records (diameter 10”) were the medium for
recording music. Long playing records turned at a constant angular
velocity of 33-1/3 rpm (revolutions per minute).
•
• What is the tangential velocity of a point on the outer edge?
•
• You have been listening to one of these recordings when you realize
that you are late for class. You lift the arm off the record and the
motor for the turntable stops. Due to the low friction bearings used
in making the turntable, the turntable does not stop rotating
instantly, but ‘coasts’ to a stop.
•
• If the turntable coasts for 12.5 complete revolutions before
stopping, calculate the constant angular deceleration and the
duration (time) of the ‘coasting’.
Mass Moment of Inertia
• The mass moment of inertia I is a mass
property of a body that measures the
resistance of the body to rotation, i.e.
• τ = Iα
• The mass moment of inertia I is similar in
rotation to the mass m in translation, i.e.
• F = ma
• Consider a point mass at a distance of r from
an axis of rotation, by definition,
• I = mr2
m
r
axis of
rotation
• If the system undergoing rotation consists of
several point masses mi at a distance of ri from
the axis of rotation, then
• I = Σ ri2 mi
• or more precisely,
• I = ∫r2dm
• The moment of inertia
of an object can change if
its shape changes. Figure
skaters who begin a spin
with arms outstretched
provide a striking example.
By pulling in their arms,
they reduce their moment
of inertia, causing them to
spin faster by the conservation
of angular momentum.
Mass Moments of Inertia for Shapes
• Thin solid disk of radius R and mass m rotating
about an axis perpendicular to the disk at its
center
• I = (1/2) mR2
• Solid cylinder of radius R, mass m about its
centerline
• I = (1/2) mR2
• Solid sphere of radius R, mass m rotating about a
diameter.
• I = (2/5)mR2
• Rod of length L and mass m rotating about an axis
normal to the rod at its center
• I = (1/12)mL2
• Rod of length L and mass m rotating about an axis
normal to the rod and at one end.
• I = (1/3)mL2
Integration to Find I
• Determine the mass moment of inertia of a
slender rod of length L and mass m with
respect to an axis that is perpendicular to the
rod and passes through one end of the rod.
• Choose a differential element between x and x
+ dx along the length of the rod.
• dm = (m/L)dx
• I = ∫x2dm = ∫x2(m/L)dx = (m/L) ∫x2dx
• I = (1/3)mL2
Class Activity
• Using integration find the mass moment of I inertia of
a slender rod of length L and mass m with respect to an
axis perpendicular to the rod at its center
• Set x= 0 at the axis of rotation, i.e. the center of the
length of the rod
• Choose a differential element between x and x + dx
along the length of the rod.
• dm = (m/L)dx
• I = ∫x2dm = ∫x2(m/L)dx = (m/L) ∫x2dx
• With the limits of integration from x = –L/2 to x = +L/2
•
•
•
•
I = (m/L) (1/3) x3 evaluated from –L/2 to +L/2.
I = (m/L)(1/3)[(L3/8) – (- L3/8)]
I = (m/L)(1/3)(L3/4)
I = (1/12)mL2
Units of I
•
•
•
•
•
•
The dimensionality of I can be found
[I] = [m][r2]
[I] = ML2
The units of I would then be
Kg m2 in the SI system
Slug ft2 in the US system