0.32 Phy1320: angular momentum -- the mathematics of torque  (Page 3/13)

Parameters for a point on a rotating disk

Consider a point on the disk that is a distance R from the axis of rotation. Assume that the disk has rotated through an angle theta. The point will havemoved through a distance, s, along the circumference of a circle of radius R. The distance s is equal to the product of the radius and the angulardisplacement measured in radians:

s = R * Q

where

• s represents the distance traveled around the circumference of a circle of radius R
• R represents the radius of the circle
• Q represents the displacement angle theta measured in radians

Relationship between speed and angular velocity

With a little calculus, we can find a relationship between the tangential speed of the point along its path around the circle and the angular velocity of the disk:

v = R*w

where

• v represents the tangential speed of the point that is moving on the circumference of the circle
• R represents the radius of the circle
• w represents the angular velocity omega

Relationship between tangential acceleration and angular acceleration

By using a little more calculus, we can find a relationship between the tangential acceleration of the point along its circular path andthe angular acceleration of the disk:

y = R*a

where

• y represents the tangential acceleration of the point along the circular path
• R represents the radius
• a represents the angular acceleration alpha

Kinetic energy in rotation

Consider the hypothetical case of a rigid object made up of a set of point particles connected by rods with zero mass. In other words, ignore the mass ofthe mechanism that holds the point particles in a rigid geometry.

Assume that this object rotates about a fixed axis with a constant angular velocity, omega. Also assume that you know the mass of each particle and that youknow the distance of each particle from the axis of rotation.

Kinetic energy of each individual particle

We know how to compute the speed of each particle along its circular path.

The kinetic energy of each particle would be

Ki = (1/2)*mi*vi^2

where

• Ki is the kinetic energy of the ith point particle
• mi is the mass of the ith point particle
• vi is the speed of the ith point particle

Kinetic energy for translational motion

Hopefully you recognize the above equation as being the same as the following equation that we saw in an earlier moduleinvolving the kinetic energy of an object in translational motion.

KE = 0.5*m*v^2

where

• KE represents the kinetic energy possessed by the object
• m represents the mass of the object
• vv represents the velocity of the object

Kinetic energy for the object

The object described above is made up of a system of particles.

The total kinetic energy of the system is the sum of the kinetic energy values of each of its particles.

The kinetic energy for the system is

Ks = (1/2)*(sum from i=0 to i=N(mi*ri^2))*w^2

where

• Ks represents the kinetic energy of the system
• ((sum from i=0 to i=N(...)) represents the sum of the values resulting from evaluating the expression in parentheses for each one of N+1 particles
• mi represents the mass of the ith point particle
• ri represents the distance of the ith point particle from the axis of rotation
• w represents the angular velocity omega