**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.

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

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