From PER wiki
Learning goals
After working through this module, you should be able to:
 Solve problems with objects rotating around a fixed axis.
 Relate rotational quantities (θ, ω and α) to linear quantities (s, v and a_{tan})
 Apply the constraints in rotational motion imposed by strings of constant length and that do not slip relative to the pulley (nonslip condition).
Applying Constrains When Solving Problems with the Rotational Form of Newton's Second Law
A common type of problems in rotational dynamics involves objects which rotational motion is constrained by the linear motion of other objects. A typical example is when different objects are connected by ropes or strings passing through pulleys.
Below we discuss the constraint imposed by a string wrapped around a massive pulley of radius R and connected to a hanging block. The resulting rotation of the pulley is related to the translation of the block because we will assume:
 Ideal string of constant length
 Nonslip condition, the string does not slip relative to the pulley.
If the string does not slip relative to the pulley, then the points of contact between the string and pulley have the same speed as the speed of a point at the edge of the pulley. In addition, if the length is constant then all the points in the string are moving at the same speed, therefore the block will move at the speed of a point at the ede of the pulley.
This is shown in the figure above. The arc length x covered by a point at the rim of the pulley (left figure) is the same as the length of the string that is displaced down by the block (right figure). Because the string does not slip relative to the pulley then x = Rθ. Taking the derivative, dx/dt = Rdθ/dt, we show that the speed of the block is the same as the speed of a point at the rim of the pulley.
In this problem, assumptions 1 and 2, imply that the speed of the block is:
and taking the derivative with respect to time the acceleration of the block is the tangential acceleration of a point at the rim:
The last relationship is particularly important, since it relates the acceleration of the block to the angular acceleration of the pulley.
The following example illustrates how the constraints of constant string length and nonslipping relative to pulley are applied to problems.
For a review of the relationship between the linear quantities (ds, v, a) of a point in a rigid object and the angular quantities (dθ, ω and α is presented at the end of this module.
Illustrative Example: Down the Well
A bucket for collecting water from a well is suspended by a rope which is wound around a pulley. The empty bucket has a mass of 2.0 kg, and the pulley is essentially a uniform cylinder of mass 3.0 kg on a frictionless axle. Suppose a person drops the bucket (from rest) into the well.
PART A: What is the bucket's acceleration as it falls?

System:

We will consider two independent systems:
 The bucket is treated as a point particle
 The pulley is treated as a rigid body.

Interactions:

1. Bucket

Resulting Forces

Point of application

bucket & Earth

m_{b}g

center of mass

bucket & rope

T: tension

center of mass

2. Pulley

Resulting Forces

Point of application

pulley & Earth

m_{p}g

center of mass

pulley & rope

T: tension

at a distance R from the center of mass

pulley & axle

F_{Axle}: contact force

center of mass

Because the mass of the pulley is uniformly distributed then the center of mass of the pulley is a the axle.


Solution

Diagramatic Representation

We begin with free body diagrams of the two two objects.

Mathematical Representation

For the bucket: we will apply Newton' second law to the bucket and write the vectors using a coordinate system with the vertical axis positive downwards, in the same direction of the acceleration:
For the pulley, there will be no translation, only rotation about the center of mass. We sum the torques about the fixed axis defined by the axle. Only the force of tension will produce a torque about the axis, therefore :
where I_{p} is the moment of inertia of the pulley about the axle, and α_{p} the pulley's angular acceleration. Note that we have followed the right hand rule to indicate the torque direction.
We now make the assumption that the rope does not stretch or slip as it unwinds. These assumptions allow us to relate the rotation rate of the pulley to the motion of the bucket:
With this assumption, we can combine the 3 equations above to obtain:
Note that is *not* equal to 1/2. The ratio of the pulley's moment of intertia to the pulley's mass times radius squared is 1/2, but we have the ratio of the pulley's moment of inertia to the bucket's mass times the pulley's radius squared.

or, using the formula for the moment of inertia of a cylinder:
As expected, the acceleration approaches g if the mass of the bucket goes to infinity, and it approaches zero if the mass of the pulley goes to infinity.




Part B: What is the bucket's speed after falling 5.0 m down the well?

System

The bucket treated as a point particle.

Interactions:

1. Bucket

Resulting Forces

bucket & Earth

m_{b}g

bucket & rope

T: tension


Solution:

Using the buckets acceleration found in part A, then we can use the expression for the function of height:
If we assume y_{i} = 0 m at the height of release, then:
The speed of the bucket, then, must be:


Review
A summary of the relationship between the linear and the angular quantities is presented below.
Relation between Linear and Angular Quantities

Consider a pulley of radius R rotating about an axis passing through its center of mass. During a time interval dt, a point B a distance r from the center will move an arc length ds. The change in the angular position dθ is related to ds by:
Based on this relationship, the linear quantities of point B, the velocity and the tangential acceleration,v and a_{tan}, are related to the angular quantities: angular speed and angular accelerations, ω and α by:
Remember that all the points on a rigid object share the same θ, ω and α although they all have different s ,v and a_{tan} . That is the reason why using angular quantities is more powerful for rigid objects

In relating the linear quantities to the angular quantities we only consider the tangential component of the acceleration. This component is the result of the change of magnitude of the velocity vector. The radial component of the acceleration is the result of changing the direction of the velocity vector and is not related to the angular acceleration.
Note: what about the radial component of the acceleration?

Recall that the acceleration of a point moving in a circle can have two components,
where the radial component is due to the change of the direction of the velocity vector:
and the tangential component is caused by a change in magnitude of the velocity vector:
If no net torque acts on a rigid object, then ω is constant and α = 0. All the points in the rigid body move in a circle with a constant ω and constant speed v.
If a point is moving with constant speed then a_{tan} = 0 but a_{r} not equal to zero.

