lecture 13: rotational kinetic energy and rotational inertia l review: rotational kinematics l...
DESCRIPTION
Comment on Axes and Signs (i.e. what is positive and negative) l Whenever we talk about rotation, it is implied that there is a rotation “axis”. l This is usually called the “z” axis (we usually omit the z subscript for simplicity). Counter-clockwise (increasing ) is usually called positive. Clockwise (decreasing ) is usually called negative. z TRANSCRIPT
Lecture 13:Lecture 13:Rotational Kinetic Energy Rotational Kinetic Energy
and Rotational Inertiaand Rotational Inertia
Review: Rotational Kinematics Rotational Kinetic Energy Rotational Inertia Torque Rotational Dynamics Equilibrium
Recall: Rotational KinematicsRecall: Rotational Kinematics
Angular Linearconstant
tαωω 0
0 021
2t t
constanta
v v at 0
x x v t at 0 021
2
And for a point at a distance R from the rotation axis:xT = RvT= RaT = R
Δθ2 220
x2a vv 220
Comment on Axes and SignsComment on Axes and Signs(i.e. what is positive and negative)(i.e. what is positive and negative)
Whenever we talk about rotation, it is implied that there is a rotation “axis”.
This is usually called the “z” axis (we usually omit the z subscript for simplicity).
Counter-clockwise (increasing ) is usuallycalled positive.
Clockwise (decreasing ) is usuallycalled negative. z
Rotational Kinetic Energy Rotational Kinetic Energy Consider a mass M on the end of a string being
spun around in a circle with radius R and angular frequency
Mass has speed v = RMass has kinetic energy
» K = ½ M v2
» K = ½ M 2 r2
Rotational Kinetic Energy is energy due to circular motion of object.
M
Kinetic Energy of Rotating DiskKinetic Energy of Rotating Disk Consider a disk with radius R and mass M,
spinning with angular frequency Each “piece” of disk has speed vi=ri
Each “piece” has kinetic energy » Ki = ½ mi v2
» = ½ mi 2 ri2
Combine all the pieces » Ki = ½ mi 2 r2 » = ½ ( mi ri
2) 2 » = ½ I 2
ri
I is called the Rotational Inertia
Rotational Inertia Rotational Inertia II Tells how much “work” is required to get object
spinning. Just like mass tells you how much “work” is required to get object moving.Ktran = ½ m v2 Linear MotionKrot = ½ I 2 Rotational Motion
I miri2 (units kg-m2)
Note! Rotational Inertia depends on what axis you are spinning about (the ri in the equation).
Rotational Inertia TableRotational Inertia Table For objects with finite number of masses, use I = m r2. For “continuous” objects, use table below:
SummarySummary
Rotational Kinetic Energy Krot = ½ I 2
Rotational Inertia I = miri2
Energy is Still Conserved!
Pulley Example Pulley Example A solid disk wheel is fixed at its center and has a
string wrapped around it and then attached to a block. The mass of the pulley is 5 kg, the radius of the pulley is 0.2 m, and the mass of the block is 8 kg. After the block has fallen 1.5 m, how fast is the block moving and how fast is the disk rotating?
We will use conservation of energy:
Remember:» Ktrans = ½ m v2
» Krot= ½ I 2
» U = m g y
5 kg
8 kg
1.5 m
ffii UKUK
fffiii mgyImvmgyImv 2222
21
21
21
21
Pulley Example Pulley Example A solid disk wheel is fixed at its center and has a string
wrapped around it and then attached to a block. The mass of the pulley is 5 kg, the radius of the pulley is 0.2 m, and the mass of the block is 8 kg. After the block has fallen 1.5 m, how fast is the block moving and how fast is the disk rotating?
Initial Krot and Ktran is zero.
Final U is zero.
5 kg
8 kg
1.5 m
22
21
21
ffi Imvmgy
Pulley Example Pulley Example A solid disk wheel is fixed at its center and has a
string wrapped around it and then attached to a block. The mass of the pulley is 5 kg, the radius of the pulley is 0.2 m, and the mass of the block is 8 kg. After the block has fallen 1.5 m, how fast is the block moving and how fast is the disk rotating?
Remember:» I = ½ m R2 for a disk » v = R so = v/R
5 kg
8 kg
1.5 m
222
21
21
21
Rv
Rmvmgym fpfbib
Pulley Example Pulley Example A solid disk wheel is fixed at its center and has a
string wrapped around it and then attached to a block. The mass of the pulley is 5 kg, the radius of the pulley is 0.2 m, and the mass of the block is 8 kg. After the block has fallen 1.5 m, how fast is the block moving and how fast is the disk rotating?
R2 cancels from the last term.
Simplify:
5 kg
8 kg
1.5 m
22
41
21
fpfbib vmvmgym
pb
bf mm
gymv
2
4
Pulley Example Pulley Example A solid disk wheel is fixed at its center and has a
string wrapped around it and then attached to a block. The mass of the pulley is 5 kg, the radius of the pulley is 0.2 m, and the mass of the block is 8 kg. After the block has fallen 1.5 m, how fast is the block moving and how fast is the disk rotating?
Solve:
5 kg
8 kg
1.5 m
vf = 4.73 m/s
TorqueTorque Rotational effect of force. Tells how effective
force is at twisting or rotating an object.
= r F sin r Fperpendicular
Units: N-mSign: CCW rotation is positive
F
r
Work Done by TorqueWork Done by Torque
Recall W = F d cos
For a wheel:W = Ftangential s
= Ftangential r ( in radians)
W = P = W/t = /t
P
EquilibriumEquilibrium Conditions for Equilibrium:
F = 0 Translational Equilibrium (Center of Mass) = 0 Rotational Equilibrium
» May choose any axis of rotation…. But Choose Wisely!
Equilibrium ExampleEquilibrium Example A 50 kg diver stands at the end of a 4.6
m diving board. Neglecting the weight of the board, what is the force each pivot exerts on the diving board?
First, draw a FBD:
F1
F2
Fg
Equilibrium ExampleEquilibrium Example A 50 kg diver stands at the end of a 4.6
m diving board. Neglecting the weight of the board, what is the force each pivot exerts on the diving board?
Write down F = 0:
-F1 + F2 – Fg = 0-F1 + F2 – (50 kg)(9.8 m/s2) =
0Note: there are two unknowns so we need another equation…
Equilibrium ExampleEquilibrium Example A 50 kg diver stands at the end of a 4.6
m diving board. Neglecting the weight of the board, what is the force each pivot exerts on the diving board?
Pick a pivot point:
F1
F2
Fg
Equilibrium ExampleEquilibrium Example A 50 kg diver stands at the end of a 4.6
m diving board. Neglecting the weight of the board, what is the force each pivot exerts on the diving board?
Write down = 0:
F1 (0 m) + F2 (1.2 m) – (50 kg)(9.8 m/s2)(4.6 m) = 0
F2 (1.2 m) – (50 kg)(9.8 m/s2)(4.6 m) = 0
Equilibrium ExampleEquilibrium Example A 50 kg diver stands at the end of a 4.6
m diving board. Neglecting the weight of the board, what is the force each pivot exerts on the diving board?
Solve:
F2 (1.2 m) – (50 kg)(9.8 m/s2)(4.6 m) = 0
F2 = 1878 N
-F1 + F2 – (50 kg)(9.8 m/s2) = 0
-F1 + (1878 N) – (50 kg)(9.8 m/s2) = 0
F1 = 1388 N
SummarySummary
Torque is a Force that causes rotation = F r sin Work done by torque: W =
Equilibrium F = 0 = 0
» May choose any axis.