Download Rotational Equilibrium and Dynamics1 Net torque: Add up individual

Rotational Equilibrium and Dynamics1
Torque: Motion of a rigid rotating object
Center of mass:



Point around which an object rotates
Not always at the center of the object
Rotating objects can have both rotational and
translational motion.
Rotational and translational motion can be separated
Ex:bowling pins spinning and flying when struck by
bowling ball
Torque (τ, tau)
τ = FdsinΘ
torque = force x length of
lever arm
SI unit: N·m (not Joule!)
Sign of torque:
Torque is a vector
+ is counterclockwise rotation
― is clockwise rotation
Net torque:
Add up individual torques,
being careful when you
assign signs for the torques.
If you are careful with signs,
the resulting sign indicates
direction of the turn.
Axis of rotation: an object is
free to rotate around a line
Example: a cat door rotating
around its hinge
Torque –
 ability of a force to rotate an object around some
axis.
 Depends on force applied and where the force is
applied (lever arm)
o The farther the force is from the axis of
rotation, the easier it is to rotate the object
and more torque is produced
o Perpendicular distance from axis of rotation
to a line drawn along the direction of the
force = lever arm
Lever arm depends on angle
Force does not have to be perpendicular for the object to
rotate – it’s just easier!
Angle < 90° requires more force!
Practice problem 7E
A basketball is being pushed by two players during tip-off.
One player exerts an upward force of 15 N at a
perpendicular distance of 14 cm from the axis of rotation.
The second player applies a downward force of 11 N at a
perpendicular distance of 7.0 cm from the axis of rotation.
Find the net torque acting on the ball about its center of
mass.
Rotational Equilibrium and Dynamics2
Simple Machines: any device that transmits or modifies force, usually by changing the
force applied to an object
6 types of simple machines:
Lever family: Lever
Pulley
Wheel and axle
Inclined plane Inclined plane
family:
Wedge
Screw
Mechanical Advantage:
how large is the output force relative to the input
force (how much does using the machine help you)
Actual Mechanical Advantage is always less than
Ideal Mechanical Advantage due to friction.
Efficiency: a measure of how well a machine works
Efficiency = Wout x 100
Win
All real machines have some friction, so efficiency
of real machines is always < 1.
MA = output force = input distance
Input force output distance
W = Fd (small force x large distance)
Or
(large force x small distance)
W=W
Rotational Equilibrium and Dynamics3
Rotation and Inertia
Determining the point around which an object rotates
Center of mass
 point around which an object rotates if gravity is the only force acting on it
 also the point where all of the mass is considered to be located in translational
motion
 Complete motion is a combination of translational and rotational motion
 In regularly-shaped objects = geometric center
 In irregularly shaped objects, the center of mass follows a parabolic path
Center of gravity
 point where gravitational force acts on the extended object as if the object were
a point mass
 for many objects, center of gravity = center of mass
Moment of inertia: resistance of an object to changes in its rotational motion; referred to as
(I)
 similar to mass (resistance to changes in motion; inertia)
 not intrinsic to the object, whereas mass is intrinsic
 Depends on mass and the distribution of that mass around the axis of rotation
o further the mass is from the axis of rotation, the greater the moment of inertia
and the more difficult it is to rotate the object
Table of Moment of Inertia values (required to solve the problems)
SI unit: kg▪m2
Rotational Equilibrium and Dynamics4
Rotational Equilibrium
When forces are equal and opposite, you expect equilibrium and the object shouldn’t
move.
But, it will rotate.
Net force = 0 (translational equilibrium)
Net torque ≠ 0 (not in rotational equilibrium)
For an object to be completely in equilibrium, both rotational and translational,
net force and net torque both must equal zero.
The absence of a net torque is the 2nd condition for equilibrium.
Steps for solving equilibrium problems:
First condition of equilibrium
1. Identify all forces
ΣF = 0
Translational equilibrium, therefore no
translational acceleration, a = 0
a. Remember to resolve into components for forces acting at an angle
2. Add up all forces acting on the object
Second condition of equilibrium
1. Choose an axis for the object to rotate around (torque).
a. The axis chosen doesn’t matter, so choose an axis that will help you! An
unknown force that acts along a line passing through the axis of rotation
produces no torque and eliminates an unknown in the problem.
b. Identify all torques
Στ = 0
Rotational equilibrium, therefore no
angular acceleration, α = 0.
i. Either the object is not moving or rotating OR it is moving and rotating
at a constant rate.
Example Problem 1:
The parents of a young girl are supporting her on a long, lightweight plank. She is sitting ¾
of the length from one parent and ¼ of the length from the other parent. Find the two
forces that support the plank and keep it from rotating. Which force is the greatest?
Rotational Equilibrium and Dynamics5
Example Problem 2:
A 5.00-m long diving board of negligible mass is supported by two pillars. One pillar is at
the left end of the diving board; the other is 1.50 m away. Find the forces exerted by the
pillars when a 90.0-kg diver stands at the far end of the board.
Sample problem:
A uniform 5.00 m long horizontal beam that weighs 315 N is attached to a wall by a pin
connection that allows the beam to rotate. Its far end is supported by a cable that makes
an angle of 53° with the horizontal, and a 545 N person is standing 1.50 m from the pin.
Find the force in the cable, FT, and the force exerted on the beam by the wall, R, if the
beam is in equilibrium.
Rotational Equilibrium and Dynamics6
Newton’s 2nd Law for Rotating Objects
Newton’s 2nd law of motion: F=ma, where acceleration depends upon mass of the object
Torque has a similar relationship with the moment of inertia.
When torque acts on an object, the resulting change in rotational motion depends on the
object’s moment of inertia.
Newton’s 2nd law for rotating objects τnet = Iα
Net torque = moment of inertia x angular acceleration
When net torque is zero, a moving object is rotating with a constant angular velocity.
Sample problem
A student tosses a dart using only the rotation of her forearm to accelerate the dart. The
forearm rotates in a vertical plane about an axis at the elbow joint. The forearm and dart
have a combined moment of inertia of 0.075 kg·m2 about the axis, and the length of the
forearm is 0.26 m. If the dart has a tangential acceleration of 45 m/s2 just before it is
released, what is the net torque on the arm and dart?
Momentum
Starting and stopping heavy object moving in a circular motion takes effort. Ex. Swinging a
sledgehammer.
 Objects resist changes in their rotational motion as well as their translational motion.
A rotating object has inertia, so it must also have momentum associated with its rotation.
This is called angular momentum.
L = Iω
Unit: kgm2
Angular
=
moment of x angular
s
Momentum
inertia
speed
When net external torque is zero, angular momentum doesn’t change. This is the basis for
the Law of Conservation of Momentum. Li = Lf
Rotational Equilibrium and Dynamics7
Sample problem
A 65 kg student is spinning on a merry-go-round that has a mass of 525 kg and a radius of
2.00 m. She walks from the edge of the merry-go-round toward the center. If the angular
speed of the merry-go-round is initially 0.20 rad/s, what is its angular speed when the
student reaches a point 0.50 m from the center?
Sample #2
A 0.11 kg mouse rides on the edge of a Lazy Susan that has a mass of 1.3 kg and a radius of
0.25 m. If the Lazy Susan begins with an angular speed of 3.0 rad/s, what is its angular speed
after the mouse walks from edge to a point 0.15 m from the center?
Rotational Equilibrium and Dynamics8
Kinetic Energy
Rotating objects also possess kinetic energy associated with their angular speed.
KErot = ½ Iω2
Unit: J
Rotational = ½ x Moment of
KE
inertia
x [angular
speed]2
Mechanical is the summation (Σ) of all types of KE and PE
ME = KEtrans + KErot + PEg
ME = ½ mv2 + ½ Iω2 + mgh
Reminder: moment of inertia still comes from the table of M of I for each type of shape.
Mechanical energy can be conserved.
MEi = MEf
Sample problem
A solid ball with a mass of 4.10 kg and a radius of 0.050 m starts from rest at a height of
2.00 m and rolls down a 30.0° slope. What is the translational speed of the ball when it
leaves the incline?