Chapter 11 Rotational Dynamics. Important quantities Moment of inertia, I. Used in =I . So, I...
-
Upload
jonathan-johnson -
Category
Documents
-
view
212 -
download
0
Transcript of Chapter 11 Rotational Dynamics. Important quantities Moment of inertia, I. Used in =I . So, I...
Chapter 11Rotational Dynamics
Important quantities
i
iiRm2
i
iivm
ii
iii
m
Rm
Moment of inertia, I. Used in =I. So, I divided by time-squared has units of force times distance (torque).
Total linear momentum is the sum of the individual momenta.
Center-of-mass is a distance. Has to have units of meters.
Rotational Relationships
2
2
1
IK
I
IL
t
t
Position
Velocity
Acceleration
Momentum
Force/Torque
Kinetic Energy
t
tt
0
20 2
1
Master equations:
One-to-one correspondence of rotational equations to linear equations.
Torque
I
mr
rrm
rma
Fr
2
Force and Torque Combined: What is the acceleration of a pulley with a non-zero moment of inertia?
R
aIITR
R
a
TR
I
Torque relation for pulley:
maTmg
Force Relation for Mass
Put it together:
2
2
1mR
Imamg
R
aIT
maTmg
21
1
mRI
ga
NOTE: Positive down!
How high does it go?Use energy conservation.
Initial Kinetic Energy:
Initial Potential Energy:
Final Kinetic Energy:
Final Potential Energy:
Zero
Zero
22
2
1
2
1 ImvK I
mgHUF
Use v=R, and then set Initial Energy = Final Energy.
2
2
2
22
12
2
1
2
1
mR
I
g
vH
mgHR
vImv
Angular Momentum
Angular momentum is “conserved” (unchanged), in the absence of an applied torque.
t
L
IL
Comparison of Linear and Angular Momentum
m
PK
t
PF
VmP
2
2
I
LK
t
L
IL
2
2
Linear momentum is conserved in absence of an applied force.
Angular momentum is conserved in absence of an applied torque.
(translational invariance of physical laws)
(rotational invariance of physical laws)
It’s time for some demos…..
L is conserved!
if
but
ffii IIL
The final moment of inertia is less, so
How about the kinetic energy?
ff
ii
I
LK
I
LK
2
22
2
Since the inertia decreases and L stays the same, the kinetic energy increases!
Q: Where does the force come from to do the work necessary to increase the kinetic energy?
A: The work is done by the person holding the weights!
Gyroscopes show change in L from applied torque
Try the bicycle wheel demo!
Static EquilibriumStatic equilibrium is achieved when both the NET FORCE and
NET TORQUE on a system of objects is ZERO.
Q: What relationship must hold between M1 and M2 for static equilibrium?
A: M1 g X1 = M2 g X2
Static balance.
What are the forces F1 and F2?
Walking the plank.
How far can the cat walk safely?
Which mass is heavier?
1. The hammer portion.2. The handle portion.3. They have the same
mass.
Balance Point
Cut at balance point
Find the Center of Mass
Left scale reads 290N, right scale reads 122N. Find total mass M and Rcm.
Find the forces.
What is the tension in the wire? What are the horizontal and vertical components of force exerted by the bolt on the rod? Let the mass of the rod be negligible.
Solution strategy has three steps:
1. Draw the free-body diagram2. Write the force equations.3. Write the torque equations.
Atwood Machine with Massive Pulley
Pulley with moment of inertia I, radius R. Given M1, M2, and H, what is the speed of M1 just before it hits the ground?
Strategy: Use conservation of mechanical energy.
Initial kinetic energy is 0. Initial potential energy is M1gH.Final kinetic energy is translational energy of both blocks plus rotational energy of pulley.Final potential energy is M2gH.Set final energy equal to initial energy.
Three steps: 1. Write down initial kinetic and potential energy.
2. Write down final kinetic and potential energy.
3. Set them equal (no friction).
HINT:
Static balance and a strange yo-yo.
The mass M of the yo-yo is known. The ratio of r and R is known. What is the tension T1 and T2, and mass m?
Strategy:
1. Write down torque equation for yo-yo.2. Write down force equation for yo-yo
and mass m.3. Eliminate unknowns.
The case of the strange yo-yo.
0
0
0
21
2
21
MgTT
mgT
RTrT
Equations to solve:
Torque
Force on mass m.
Force on yo-yo.
mgTr
RTT
2
21
Mg
M is known, R/r is known.
Step 1: Rearrange. Step 2: Substitute.
Mgr
RT
MgTr
RT
1
0
2
22
1
1
1
1
2
a
Mm
a
aMgT
a
MgT
Step 3: Solve (a=R/r).