1. Physics 106: Mechanics Lecture 03
Wenda Cao
NJIT Physics Department

2. Rotational Equilibrium and Rotational Dynamics II
Rotational Kinetic Energy
Moment of Inertia
Torque
Newton 2nd Law for Rotational Motion: Torque and angular acceleration
Februaryl 3, 2011

3. Rotational Kinetic Energy
There is an analogy between the kinetic energies associated with linear motion (K = ½ mv 2) and the kinetic energy associated with rotational motion (KR= ½ Iw2)
Rotational kinetic energy is not a new type of energy, the form is different because it is applied to a rotating object
Units of rotational kinetic energy are Joules (J)
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4. Moment of Inertia of Point Mass
For a single particle, the definition of moment of inertia is
m is the mass of the single particle
r is the rotational radius
SI units of moment of inertia are kg.m2
Moment of inertia and mass of an object are different quantities
It depends on both the quantity of matter and its distribution (through the r2 term)
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5. Moment of Inertia of Point Mass
For a composite particle, the definition of moment of inertia is
mi is the mass of the ith single particle
ri is the rotational radius of ith particle
SI units of moment of inertia are kg.m2 cm m P L cm m P
L/2 cm m P L
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6. Moment of Inertia of Extended Objects
Divided the extended objects into many small volume elements, each of mass Dmi
We can rewrite the expression for I in terms of Dm
With the small volume segment assumption,
If r is constant, the integral can be evaluated with known geometry, otherwise its variation with position must be known
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7. Moment of Inertia for some other common shapes
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8. Februaryl 3, 2011

9. Parallel-Axis Theorem
In the previous examples, the axis of rotation coincided with the axis of symmetry of the object
For an arbitrary axis, the parallel-axis theorem often simplifies calculations
The theorem states
I = ICM + MD 2
I is about any axis parallel to the axis through the center of mass of the object
ICM is about the axis through the center of mass
D is the distance from the center of mass axis to the arbitrary axis
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10. Rotation axes perpendicular to plane of figure
Masses on the corners of a rectangle, sides a & b
h2 = (a/2)2 + (b/2)2 X cm b a h m P
About an axis through the CM:
About an axis “P” through a corner:
Using the Parallel Axis Theorem directly for the same corner axis:
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11. Force vs. Torque Forces cause accelerations
What cause angular accelerations ?
A door is free to rotate about an axis through O
There are three factors that determine the effectiveness of the force in opening the door:
The magnitude of the force
The position of the application of the force
The angle at which the force is applied
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12. General Definition of Torque
Let F be a force acting on an object, and let r be a position vector from a rotational center to the point of application of the force. The magnitude of the torque is given by
 = 0° or  = 180 °:
torque are equal to zero
 = 90° or  = 270 °: torque attain to the maximum
Torque will have direction
If the turning tendency of the force is counterclockwise, the torque will be positive
If the turning tendency is clockwise, the torque will be negative
The applied force is not always perpendicular to the position vector r. Suppose the force F is directed away from the axis, say, I grasp the doorknob and pushing to the left. Can I open the door? Here, you can see although a force is acted on the door, angular acceleration is equal to zero. However, if the applied force acts at an angle to the door, the component of the force perpendicular to the door will cause it to rotate. This phenomena imply that the torque must be related with the angle at which the force is applied. These considerations motivate a more general definition of torque: Let F be a force acting on an object, and let r be a position vector from a rotational center to the point of application of the force. The magnitude of the torque is given by …
Februaryl 3, 2011

13. Net Torque
The force will tend to cause a counterclockwise rotation about O
The force will tend to cause a clockwise rotation about O
St = t1 + t2 = F1d1 – F2d2
If St  0, starts rotating
If St = 0, rotation rate does not change
Rate of rotation of an object does not change, unless the object is acted on by a net torque
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14. Torque on a Rotating Object
Consider a particle of mass m rotating in a circle of radius r under the influence of tangential force
The tangential force provides a tangential acceleration: Ft = mat
Multiply both side by r, then
rFt = mrat
Since at = r, we have
rFt = mr2
So, we can rewrite it as
 = mr2
 = I
Februaryl 3, 2011

15. Torque on a Solid Disk Consider a solid disk rotating about its axis.
The disk consists of many particles at various distance from the axis of rotation. The torque on each one is given by
 = mr2
The net torque on the disk is given by
 = (mr2)
A constant of proportionality is the moment of inertia,
I = mr2 = m1r12 + m2r22 + m3r32 + …
So, we can rewrite it as
 = I
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16. Newton’s Second Law for a Rotating Object
When a rigid object is subject to a net torque (≠0), it undergoes an angular acceleration
The angular acceleration is directly proportional to the net torque
The angular acceleration is inversely proportional to the moment of inertia of the object
The relationship is analogous to
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17. Februaryl 3, 2011

18. Example 1: second law for rotation
When she is launched from a springboard, a diver's angular speed about her center of mass changes from zero to 6.20 rad/s in ms. Her rotational inertia about her center of mass is constant at kg·m2. During the launch, what are the magnitudes of
(a) her average angular acceleration and
(b) the average external torque on her from the board?
a) Use: or
b) Use:
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19. Example 2: a for an unbalanced bar
m1g
m2g L2 L1
fulcrum +y
a1 = - aL1
a2 = + aL2
Constraints:
Bar is massless and originally horizontal
Rotation axis at fulcrum point
 N has zero torque
Find angular acceleration of bar and the linear
acceleration of m1 just after you let go
Use:
Using specific numbers:
where:
Let m1 = m2= m
L1=20 cm, L2 = 80 cm
What happened to sin(q) in moment arm?
net torque
Clockwise
Accelerates UP
total I about pivot
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20. Newton 2nd Law in Rotation
Suppose everything is as it was in the preceding example, but the bar is NOT horizontal. Assume both masses are equal. Which of the following is the correct equation for the angular acceleration? N
m2g L2 L1
fulcrum
m1g q
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21. Strategy to use the Newton 2nd Law
Many components in the system means several (N) unknowns….
… need an equal number of independent equations
Draw or sketch system. Adopt coordinates, name the variables, indicate
rotation axes, list the known and unknown quantities, …
Draw free body diagrams of key parts. Show forces at their points of
application. find torques about a (common) axis
May need to apply Second Law twice to each part
Translation:
Rotation:
Make sure there are enough (N) equations; there may be constraint
equations (extra conditions connecting unknowns)
Simplify and solve the set of (simultaneous) equations.
Interpret the final formulas. Do they make intuitive sense? Refer back
to the sketches and original problem
Calculate numerical results, and sanity check anwers (e.g., right order of
magnitude?)
Note: can have
Fnet .eq. 0
but tnet .ne. 0
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22. Rotating Rod
A uniform rod of length L and mass M is attached at one end to a frictionless pivot and is free to rotate about the pivot in the vertical plane as in Figure. The rod is released from rest in the horizontal position. What are the initial angular acceleration of the rod and the initial translational acceleration of its right end?
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23. Rotating Rod
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24. The Falling Object
A solid, frictionless cylindrical reel of mass M = 2.5 kg and radius R = 0.2 m is used to draw water from a well. A bucket of mass m = 1.2 kg is attached to a cord that is wrapped around the cylinder.
(a) Find the tension T in the cord and acceleration a of the object.
(b) If the object starts from rest at the top of the well and falls for 3.0 s before hitting the water, how far does it fall ?
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25. Example, Newton’s Second Law for Rotation
Draw free body diagrams of each object
Only the cylinder is rotating, so apply St = I a
The bucket is falling, but not rotating, so apply SF = m a
Remember that a = a r and solve the resulting equations
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26. Cord wrapped around disk, hanging weight
Cord does not slip or stretch  constraint
Disk’s rotational inertia slows accelerations
Let m = 1.2 kg, M = 2.5 kg, r =0.2 m r a mg
For mass m: T mg y
Unknowns: T, a
support force
at axis “O” has
zero torque
FBD for disk, with axis at “o”: N Mg T
Unknowns: a, a
from “no slipping” assumption
So far: 2 Equations, 3 unknowns Need a constraint:
Substitute and solve:
Februaryl 3, 2011

27. Cord wrapped around disk, hanging weight
Cord does not slip or stretch  constraint
Disk’s rotational inertia slows accelerations
Let m = 1.2 kg, M = 2.5 kg, r =0.2 m r a mg
For mass m: T mg y
Unknowns: T, a
support force
at axis “O” has
zero torque
Februaryl 3, 2011