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PH 121 12.5-12.7

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by

Richard Datwyler

on 10 December 2013

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Transcript of PH 121 12.5-12.7

Torque
Welcome back.
Today we are going to take this rotational dynamics and
bring it up to speed with all the concepts of motion.

Starting specifically with the application of forces.

The ability of a force to cause rotation depends on
three things:
1. The magnitude of the force.
2. The distance the force is from the pivot.
3. The angle of the force.

Consider opening a door.
How do these things affect the ability I have
to open this door?
As noted the torque is defined as:


The only tricky bit is the angle.
It is measured between the vectors.
With that, we have defined how
a force causes rotation.

Torque then takes the place of all
'forces' in our rotational dynamics.

Like other rotational variables it
has a sign that is positive when
causing CCW rotation, and negative
with CW rotation.
As we consider this equation of torque
we can view it in two different ways.
The 'sine' can go with the Force
or the distance.

If with the force then it implies the
tangential component of the force.
If with the radial distance then it is
called the moment arm.




Both ideas are just fine.
F
r
d
Just as with forces acting on a particle
we need to add up all the torques
acting on a rigid body.
a few good steps to consider
1. Identify the 'forces'
2. Cleverly pick a pivot or axis of rotation
3. Note the radial distance from axis to force
4. Note the angle between them.
5. Assign positive or negative values.
As an example here is a beam with two
masses on the ends.
A
B
C
D
Which of these forces
will produce the most
Torque
A
B
C
D
At which location
would the axis be
best located
A
B
C
D
Which of these forces
will produce a positive
value of torque?
A A & B
B C & D
C B
D A
As you may have noticed there was a bit of
a random force acting on this rigid body at
point C.

This is a special torque of gravity.

Gravity is acting at all locations on this beam
And we could add up all of these individual
torques and find a resultant




This says that the Torque due to gravity
can be considered by taking the force of gravity
acting on the objects Center of mass only
Dynamics
F
F
F
t
r
Consider the force
shown here acting on
the rigid body about this
pivot
Only the tangential force
will cause it to move
and the only acceleration
will be in the
tangential direction
If we look at this force and use Newton's
2nd law we have
Using the definition of alpha we have:
Multiply both sides by r
But that is one definition of Torque
Finally if we summed up all of these
we have a moment of inertial
This then is Newton's
2nd law for rotations
Before we jump into solving
some problems
Let us consider a constraint
to the system.

If a disk is to be rotated, by having a
rope pulled over the outside edge
'think rope and pulley'
we assume they don't slip, and thus
they are constrained to move together

The rope produces a Torque, and the
angular velocity/acceleration of the pulley
and the
tangential velocity/acceleration of the
rope are fixed

When a circle 'rolls' down a hill, what is causing that
rolling motion?
"When measuring torque on a bolt, where is the measurement taking place at? "
"So torque is strictly for rotating objects?"
"Can we go over example 12.13?"
"Could you please explain the relationship between torque and angular acceleration?"
"could you just explain the relation between newtons second law and torque?"
"Gravitational Torque, not sure how to find it?"
An object's moment of inertia is 2.0 kg m^2. Its angular velocity is increasing at a rate of 4.0 rad/s per second. What is the torque on the object?
8 Nm
Blocks of mass m1 and m2 are connected by a massless string that passes over a pulley. The pulley turns on frictionless bearings. Mass m1 slides on a horizontal, frictionless surface. Mass m2 is released while the blocks are at rest.

A. Assume the pulley is massless. Find the acceleration of m1 and the tension in the string
B. Suppose the pulley has a mass of mp and a radius of R. Find the acceleration of m1 and the tensions in both parts of the string.
m 1
m 2
The 2.0 kg, 30 cm diameter disk is spinning at 300 rpm. how much friction force must the brake apply to the rim to bring the disk to a halt in 3.0 s?
4.7 N
example 12.13
Full transcript