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ROTATIONAL DYNAMICS
CHAPTER 12
12-1 Rotational Dynamics: An Overview
12-2 Kinetic Energy of Rotation and Rotational Inertia
12-3 Rotational Inertia of Solid Bodies 12-4 Torque Acting on a Particle
12-5 Rotational Dynamics of a Rigid Body
Units of Chapter 12
• Forces cause accelerations (f = ma)
• 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
• The quantity takes into accounts these factors Called
Torque (t)12-1 Rotational Dynamics An Overview
Force vs. Torque
General Definition of Torque
• The applied force is not always perpendicular to the position vector
• When the force is parallel to the position vector, no rotation occurs
• When the force is at some angle, the perpendicular component causes the rotation
Torque, t , is the tendency of a force to rotate an object about
some axis
• An object rotating about z axis with an angular speed ω, has rotational kinetic energy
• The total rotational kinetic energy of the rigid object is the sum of energies of all its particles
• Where I is called the rotational inertia 𝐼 = 𝑚
𝑖𝑟
2𝑖• The unit of rotational kinetic energy is Joule (J)
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2 2
2 2 2
1 2
1 1
2 2
R i i i
i i
R i i
i
K K m r
K m r I
12-2 Kinetic Energy of Rotation and Rotation
Inertia
Parallel axis theorem
• The rotational of inertia of any body about an arbitrary axis equal the rotational of inertia about a parallel axis through the center of mass plus the total mass times the squared distance between the two axes.
Where I is the rotational inertia about the arbitrary axis, I
cmIs the
rotational inertia about the parallel axis through the Center of mass,
and h is the distance between the axes.
Sample problem 2 (p248)
Sample Problem 2: The Object shown in fig. consists of two particles, of masses m
1and m
2, connected by light rigid rod of Length L. (a) Neglecting the mass of rod, find the rotational inertia I of this system for rotation of this object about an axis perpendicular to the rod and a distance x from m
1. (b) Show that I is minimum when x=x
cmSolving, we find the value of x at which this minimum occurs:
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We can evaluate the Rotational Inertia of an extended rigid object by imagining the object divided into many small volume elements, each of which has mass δm.
As in The figure the rod has length L and mass M.
let us divided the rod into 10 pieces
For 20 pieces
12-3 Rotational Inertia of Solid Bodies
for any arbitrary number N of pieces:
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𝑰 = 𝒓𝒊𝟐 𝜹𝒎𝒊
In this limit, the sum becomes an integral over the whole object:
𝑰 = 𝒍𝒊𝒎
𝜹𝒎→𝟎 𝒓𝒊𝟐 𝜹𝒎𝒊 = 𝒓𝟐 𝒅𝒎 In general we use the definition
As an example, let us return to the rod rotated about an axis through its center ρ= m /V,
We have where ρ is the density of the object and V is its volume
ρ= dm / dV dm = ρ dV dm = ρ dV = ρ A dx
𝑰 = 𝒍𝒊𝒎
𝜹𝒎→𝟎 𝒓𝒊𝟐 𝜹𝒎𝒊 δm→0
Some Rotational Inertias
• 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, with F perpendicular to r, the torque is given by the vector (or cross product)
F r
t
12-4 Torque Acting on a Particle
The SI units of torque are N.m, Torque is a vector quantity
Torque magnitude is given by
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
• 0° or 180 °: torque are equal to zero
• 90° or 270 °: magnitude of torque attain to the maximum
t rF sin
Understand 𝒔𝒊𝒏 𝜽
The moment arm, d, is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force d = r sin
Fd rF
t sin
The component of the force ( F cos ) has no tendency to produce a rotation
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Sample Problem 3: A pendulum consists of a body of mass w =
0.17 kg on the end of a rigid rod of length L= 1.25 m and negligible mass (see Fig.). a)What is the magnitude of the torque due to
gravity about the pivot point O at the instant the pendulum is
displaced as shown through an angle of 𝜃 = 10𝑜 from the vertical?
(b) What is the direction of the torque about O at that instant?
Does its direction depend on whether the pendulum is displaced to the left or right of vertical?
Sample Problem 3
(b) With the displacement as shown in Fig. , the torque the torque about the pivot is into the plane of the paper.
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12.5 Rotational Dynamics of a Rigid Body
The point p, which is at a distance r from the axis of rotation, moves through a distance ds = rdφ as the body rotates through the angle (dφ). The work (dW) can therefore be expressed as
t
d rd
F
dW
dW t d
During the interval dt, the work-energy theorem gives
t
W
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t I
• This equation is the rotational analogue of Newton's second law in the scalar form.
• When a rigid object is subject to a net torque (𝜏), it undergoes an angular acceleration
• The angular acceleration is directly proportional to the net torque
• The angular acceleration is inversely proportional to the Rotational of inertia of the object
The mechanical power
t
t
dt
d dt
P dW
t d
dW P t
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COMPARISON OF LINEAR AND ROTATIONAL IWNAMICAL
EQUATIONS
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Calculate the rotational inertia of a meter stick, with mass 0.56 kg, about an axis perpendicular to the stick and located at the 20 cm mark.
We use the parallel axis theorem:
I = Icm + Mh2 ,
where Icm is the rotational inertia about the center of mass, M is the mass, and h is the distance between the center of mass and the chosen rotation axis. The center of mass is at the center of the meter stick, which implies
h = 0.50 m – 0.20 m = 0.30 m.
We find
Solution
