Transcript
Chapter 11: Rotational Dynamics and Static Equilibrium
Torque: The ability of a force to rotate a body
about some axis. rF Note: F r
The torque is larger if the force is applied further from the axis of rotation.
Rotational Dynamics• Which force has the greatest
effect on the rotation of the door?
a
b
c
pivot
By convention, the sign of torque is:
< 0 clockwise (cw)
> 0 counter-clockwise (ccw)
General Definition of TorqueOnly the component of the force that is perpendicular to the radius causes a torque.
= r(Fsin)
Equivalently, only the perpendicular distance between the line of force and the axis of rotation, known as the moment arm r, can be used to calculate the torque.
= rF = (rsin)F
F
Fsin
Each force that acts on an object may cause a torque.F1
When discussing torques, we must identify a pivot point (or axis of rotation).
In this figure, the three forces have equal magnitude.
• Which forces cause a torque?
• Which force causes the biggest magnitude torque?
• Which forces, if any, causes a positive torque?
r1r2
F2
F3
pivot point
The net torque about a point O is the sum of all torques about O:
Example 1
Calculate the net torque on the 0.6-m rod about the nail at the left. Three forces are acting on the rod as shown in the diagram.
30°
0.3 m
4 N
5 N
6 N
Moment of InertiaRecall that mass (inertia) is an object’s resistance to
acceleration. Similarly an object’s resistance to rotation (angular acceleration) is known as moment of inertia. For a point mass m:
I = mr2
I = moment of inertiar = distance from the axis of rotationFor an extended object:
I =miri2
Mass near the axis of rotation resists rotation less than mass far from the axis of rotation.
For circular motion, the distance (arc length) s, the radius r, and the angle are related by:
r
tst
)()(
DEGRAD 180
Note that is measured in radians:
Angular Position,
> 0 for counterclockwise rotation from reference line
1 rev = 360° = 2 rad
Consider a rotating disk:
t = 0
O P
r
t > 0
O
Pr s
Notice that as the disk rotates, changes. We define the angular displacement, , as:
= f - i
which leads to the average angular speed av
if
ifav ttt
Angular Velocity,
Instantaneous Angular Velocity
As usual, we can define the instantaneous angular velocity as:
tt
lim0
Note that the SI units of are: rad/s = s-1
> 0 for counterclockwise rotation< 0 for clockwise rotation
If v = speed of a an object traveling around a circle of radius r
= v / r
Walker Problem 4, pg. 297Express the angular velocity of the second hand on a clock in the following units: (a) rev/hr (b) deg/min and (c) rad/s.
How long does it take for the second hand to complete one revolution?How many degrees in one revolution?How many radians in one revolution?
PeriodThe period of rotation is the time it takes to complete one revolution.
T
2 T = period
Rearranging we have2
T
What is the period of the Earth’s rotation about its own axis?
What is the angular velocity of the Earth’s rotation about its own axis?
We can also define the average angular acceleration av:
and
if
ifav ttt
tt
lim0
Angular Acceleration,
The SI units of are: rad/s2 = s-2
We will skip any detailed discussion of angular acceleration, except to note that angular acceleration is the time rate of change of angular velocity
Torque and Angular Acceleration
Recall Newton’s Second Law: F = ma
The net force on an object of mass m causes a (linear) acceleration a.
Similarly, the net torque on an object with moment of inertia I causes an angular acceleration .
= I
Partial derivation of role of torque.
• Consider a point mass m constrained to
move in a circle of radius r.
• The centripetal acceleration is ac = v2/r
• The change in speed is determined by the component of force parallel to motion (and therefore perpendicular to the radius): ma|| = F = Fsin mr = F
= rF / mr2 = / I
In order for a system to not change its state of rotation, the external torque applied to the system must be 0.
F
r
v
Consider the wheel shown below. Two forces of equal magnitude are acting on the wheel. Will the wheel remain at rest? NO – it will rotate!
The net force is zero, so there will be no linear acceleration.
Zero Torque and Static Equilibrium
However, the sum of the torques is not zero, so there will be an angular acceleration.
The wheel is not in static equilibrium.
F1
F2
Conditions for Static Equilibrium
For true static equilibrium, two conditions must be satisfied:
For an object in equilibrium, the axis of rotation is arbitrary (But all torques must be evaluated about a common axis).
0
0
F 0
0
y
x
F
F
Walker Problem 30, pg 341A rigid, vertical rod of negligible mass is connected to the floor by an axle through its lower end, as shown in the Figure. The rod also has a wire connected between its top and the floor. If a horizontal force F is applied at the midpoint of the rod, find (a) the tension in the wire, and (b) the horizontal and vertical components of force exerted by the bolt on the rod.
FT
mg0
P
Static Equilibrium:Fx = 0 F +Tcos(180º+45º) + Px = 0fy = 0 Py+Tsin(180º+45º)-mg = 0 = 0 (F)(L/2) +(Tcos45º)L = 0
x
y
T = F / (2 cos45º) = F / 2cos(180º+45º) = 1/ 2 = sin (180º+45º) Px = F – Tcos(180º+45º) = F – (F/2)(-1/ 2)Px = F + F/2 = F/2Py = mg + T/ 2 = mg + F/2 F/2
Center of Mass and BalanceRecall that an object will hang with it center of mass (CM) directly below the point of suspension. Now we can understand why. If the object’s CM is not below the point of suspension, its weight will cause a torque which rotates the object until its CM is below the point of suspension.
(a) gravitational torque = 0
(b)gravitational torque rotates the paintbrush
Base of Support
An object at rest on a surface is in equilibrium (will not tip over) if its center of mass is above the base of support. The base of support is the area bounded by whatever is touching the floor.
The shaded area between the legs of the table is the base of support.
Why isn’t it possible to touch your toes if you are standing flush against a wall?
Walker Problem 38, pg. 342
A baseball bat balances 71.1 cm from one end. If a 0.560-kg glove is attached to that end, the balance point moves 24.7 cm toward the glove. Find the mass of the bat.
Angular Momentum
For linear momentum:
p = mv
For rotational motion, we define an angular momentum:
L = r mv = mr2 = I
The SI units of angular momentum are kg·m2/s
Walker Problem 54, pg. 343
Two gerbils run in place with a linear speed of 0.45 m/s on an exercise wheel that is shaped like a hoop. Find the angular momentum of the system if each gerbil has a mass of 0.33 kg and the exercise wheel has a radius of 9.5 cm and a mass of 5.0 g.
V=0.45 m/s
All mass is at radius r = 0.095mI = mr2 = (0.005 + 2*0.33)kg (0.095m)2
I = 0.00600 kg m2
L = I = v/r = (0.45 m/s) / (0.095 m) = 4.74/sL = (0.00600 kg m2) (4.74 /s)L = 0.028 kg m2/s
Recall that Ft = p
For rotational motion:
t = L
Conservation of Angular MomentumIf then 0 fi LL
With zero external torque, Angular momentum is constant, even if internal forces cause a change in the distribution of mass.
• Elliptic orbits (chap. 12): L = constant• Ice skater moving arms in, radius shrinks, Moment of inertia
I shrinks, L = I = constant, w increases.
Walker Problem 61, pg. 343A student sits at rest on a piano stool that can rotate without friction. The moment of inertia of the student-stool system is 4.1 kg·m2. A second student tosses a 1.5-kg mass with a speed of 2.7 m/s to the student on the stool, who catches it at a distance of 0.40 m from the axis of rotation. What is the resulting angular speed of the student and the stool? Assume that the velocity of the mass, before it is caught, is tangential to a circle of radius 0.4 m from the axis of rotation.
Walker Problem 63, pg. 343A turntable with a moment of inertia of 5.4 10-3 kg·m2 rotates freely with an angular speed of 33.33 rpm. Riding on the rim of the turntable, 15 cm from the center, is a 1.3-g cricket. (a) If the cricket walks to the center of the turntable, will the turntable rotate faster, slower, or at the same rate? Explain. (b) Calculate the angular speed of the turntable when the cricket reaches the center.
Kinetic energy of rotation
What is the kinetic energy of a mass m traveling at speed v in a circle of radius r?K = (1/2) m v2 = (1/2) mr2 (v/r) 2 = (1/2) I 2
Kinetic energy of rotation = (1/2) I 2 = L2 / (2I)
This is not a new form of energy, just a re-labeling (or alternate formula) for kinetic energy.
Quiz• Two masses, 1kg and 0.5 kg,are balanced
on a fulcrum at radii 0.20m and 0.40 m, respectively.
• Now the lighter mass is hung from the same position as before, but with a string of length 0.20 m.
• Choose the correct answer:
1. The 0.5 kg mass swings down (longer length = larger torque)
2. The system remains balanced (torques don’t change)
3. The 0.5 kg mass swings up (it is lighter than the 1kg mass).
1kg 0.5kg
1kg
0.5kg
Gravitational Torque
• Gravity generates a torque as if the entire force were concentrated at the Center-of-Mass.
• Consider a discrete set of point masses m1, m2,
• Gravitational torque acting on each mass is x-coordinate times force of gravity (since gravity is perpendicular to x-direction)
• Torque = x1 m1 g + x2 m2 g + …
• = (x1 m1 + x2 m2 )g = XCM Mg
m1gm3g
x
y
Gravitational Potential Energy
• Single mass U = mgh
• Mass distribution U = mgyCM
Stability
• An object in equilibrium is stable, if any tiny displacement (rotation or translation) causes the center of gravity to rise.
Gravitational Stability
• Stable– CM rises
• Unstable– CM falls
Angular Momentum
For linear momentum:
p = mv
For rotational motion, we define an angular momentum:
L = r mv = mr2 = I
The SI units of angular momentum are kg·m2/s
Recall that Ft = p
For rotational motion:
t = L
Conservation of Angular MomentumIf then 0 fi LL
With zero external torque, Angular momentum is constant, even if internal forces cause a change in the distribution of mass.
• Elliptic orbits (chap. 12): L = constant• Ice skater moving arms in, radius shrinks, Moment of inertia
I shrinks, L = I = constant, w increases.
Kinetic energy of rotation
What is the kinetic energy of a mass m traveling at speed v in a circle of radius r?K = (1/2) m v2 = (1/2) mr2 (v/r) 2 = (1/2) I 2
Kinetic energy of rotation = (1/2) I 2 = L2 / (2I)
This is not a new form of energy, just a re-labeling (or alternate formula) for kinetic energy.
Hurricane
• Coriolis Force = Conservation of Angular Momentum
• Pressure differential• (1000mBar outside-
882 mBar inside Wilma)
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