# Ch 10 Dynamics of Rotational Motion-final .pdf

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University of Toronto Mississauga

Physics

PHY136H5

Wagih Ghobriel

Fall

Description

cHaPtER 10
Rotational
Dynamics According to Newton’s second law,
a net force causes an
object to have an acceleration.
What causes an object to have an
angular acceleration? TORQUE The amount of torque depends on
where and in what direction the
force is applied, as well as the
location of the axis of rotation. Definition of Torque
Magnitude of Torque = (Magnitude of the force) x (Lever arm)
τ = F
Torque (τ) is defined as the force applied multiplied by the moment arm (or lever arm).
Direction: The torque is positive when the force tends to produce a
counterclockwise rotation about the axis, and negative when the force tends to
produce a clockwise rotation.
SI Unit of Torque: newton.meter (N·m) Why do acrobats carry long bars?
The acrobat controls the net torque acting on her center of gravity by manipulating the pole.Also,
the acrobat holds the pole low, moving her center of gravity closer to the rope. Sign Convention
A counterclockwise
torque is designated as
positive(+)
A clockwise torque
is designated as
negative (─) Why internal forces do not
contribute to net torque Example:
A plumbing problem to solve
Example 10.1
• Refer to the worked problem on page 296. Example The Achilles Tendon
The tendon exerts a force of magnitude
790 N. Determine the torque (magnitude
and direction) of this force about the
ankle joint.
τ = F
790 N
cos55 = 3.6 10 m
−2
τ = 720 N 3.6×10 m cos55
=15 N⋅m Center of Gravity (Revisited)
Definition of
Center of Gravity
The center of gravity of a
rigid body is the point at
which its weight can be
considered to act when the
torque due to the weight is
being calculated.
Suppose we have a group of objects, with
known weights and centers of gravity, and
it is necessary to know the cg fWhen an object has a symmetrical
group as a whole: shape and its weight is distributed
uniformly, the center of gravity lies at
Wx1 1 x + 2 2 ... its geometrical center.
xcg
W 1 + W 2 + ... Applying an external force by the index
finger to provide balance
Horizontal board
and a weight Horizontal board
W x +W x +
x = 1 1 2 2
cg
W 1W + 2 Example
The Center of Gravity of an Arm
The horizontal arm is composed
of three parts: the upper arm
(17 N), the lower arm (11 N),
and the hand (4.2 N).
Find the center of gravity of the
arm relative to the shoulder
joint.
x = W1 1+W 2 2+
cg W 1+W 2 +
17 N 0.13m + 11N 0.38m + 4.2 N 0.61m )
xcg= = 0.28m
17 N+11N+4.2 N Conceptual Example Overloading a Cargo Plane
This accident occurred because the plane was overloaded toward the rear. How did a
shift in the center of gravity of the plane cause the accident? Finding the center of gravity of an irregular shape. 10.2 Torque and Angular Acceleration
Newton’s Second Law for Rotational
Motion About a Fixed Axis
F = ma
T T
τ = FTr aT= rα
τ = mr α 2
I = mr is the moment of
inertia of the body
Notes 2
N2 is valid only in an inertial frame.
Net torque of the internal forces cancel.
Although a rigid object possesses a unique total mass, it does not have a unique
moment of inertia. 2
2 τ1= m 1 1α
∑ τ = ∑ (mr α 2
τ2= m 2 2α
Net external
torque Moment of
inertia τN= m N Nα Rotational analog of newton’s second law
for a rigid body rotating about a fixed axis
Moment of Angular
Net externaltorque=inertia acceleration
Requirement:Angular acceleration
∑ τ = I α must be expressed in radians/s .
I= (mr 2
∑
Example: Rotating masses
With respect to the variables given in the figure, the equation for the angular acceleration α is:
(a) rF/2 mR (b) RF/2 mr (c) rF/3 mR (d) RF/3 mr (e)2 /3 mr Reminder
Example The Moment of Inertial Depends on Where the Axis Is.
Two particles each have mass m and are fixed at the ends of a thin rigid rod.
The length of the rod is L. Find the moment of inertia when this object rotates
relative to an axis that is perpendicular to the rod at (a) one end and (b) the center.
(a) I = ∑ mr = m r 1 1r = 2 2 +m L )2 ( )2 2
I = mL
m 1 m =2m r = 0 r = L
1 2
(b) I = ∑ (mr = m r +m r = m L 2 +m L 22 ( )2 1 2
1 1 2 2 I = m2
r = L 2 r = L 2
1 2 Reminder Reminder Example: Hoisting a Crate
The combined moment of inertia of the dual pulley is 50.0 kg·m 2. The crate weighs
4420 N. Atension of 2150 N is maintained in the cable attached to the motor. Find
the angular acceleration of the dual pulley. equal
F = T −mg = ma ∑ τ =T1 1 2 2α
∑ y 2 y
ay= 2α T 2 mg +ma y T1 1− mg m+ y)2 = Iα
ay = 2α
T − mg +m α = Iα
1 1 2 2
T −mg
α = 1 1 2 2
I +m 2
(2150 N 0.600m − 451kg 9.80m s 2)(0.200m ) 2
= 2 2 = 6.3rad s
46.0kg m + 451kg 0.200m ) The
yo-yo
rotates
on a
moving
axis. A bowling ball rotates on a moving axis.
Refer to worked example
on pages 302 and 303. 10.3 Rotational Work, Energy and Power
in Rotational Motion
W = Fs = Fr θ
s = rθ τ = Fr
W =τθ
Definition of Rotational Work
The rotational work done by a constant torque
in turning an object through an angleΔθ is
W R = τ.
Requirement:
SI Unit of Rotational Work:joule (J)s. Work can be done by
a constant torque.
Example 10.6 follows the
generic situation to the right. 2 2 2
KE = 2v =Tmr 2
v = rω
T
1 2 2 1 2 2 1 2
KE = ∑ (2mr ω = ) (2 ∑ mr ω = Iω 2
Definition of Rotational Kinetic Energy
The rotational kinetic energy of a rigid rotating object is
2
KE =RIω2
Requirement:
The angular speed must be expressed in rad/s.
SI Unit of Rotational Kinetic Energy:joule (J)
Power in Rotationalal Motion
∆∆ θ
P = = τ = τω
∆∆t 10.4 Angular Momentum
Definition of Angular Momentum
The angular momentum L of a body rotating about a fixed axis is the
product of the body’s moment of inertia and its angular velocity with
respect to that axis:
L = Iω
Requirement:
The angular speed must be expressed in rad/s.
SI Unit of Angular Momentum:kg·m
Newton’s second law
of motion Conservation
Impulse-momentum Sum of average of angular
Impulse of a torque. t

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