Physics 211 Rolling Motion of a Rigid Body Vector Product and Torque Angular Momentum Rotation of a Rigid Body about a Fixed Axis Conservation of Angular Momentum 10: Angular Momentum and Torque

Rolling Motion of a Rigid Body Rotation + Translation  If v cm  r    Pure Rolling Motion

As then one can see that the center of mass moves at v cm by noting that  2  f  angular frequency= angular speed  where f  frequency of rotation, which has units rps = revolutions per second or cps = cycles per second Which shows that the distance traveled by the center of mass in pure rolling motion in one second is 2  r  f  2  r   2   r 

v cm v top = v cm + v v bottom = v cm - v Pure rolling motion = no skidding v = linear speed of rim of wheel just due to the rotational motion; v top & v bottom are the total linear speeds at the top and bottom

v = v cm thus for pure rolling motion v top = v cm + v cm = 2v cm v bottom = v cm - v cm = 0!!!!!!

Skidding = no rotational motion about of center of mass

N W FsFs No work done by non conservative forces in pure rolling motion  tot.mech  =   K tot  = -  U tot h   U tot = -mgh   K tot = K tot,final - K tot,initial = K tot,final (as initially at rest)  K tot = K rot + K trans

 As there is no slipping (skidding) the rolling object only experiences STATIC friction with the surface  Static friction can NEVER do any work  At each instant of time the static friction stops translational motion and causes rotation  [The static friction does not have to be the maximum possible static friction (i.e. it can be ]  The static friction produces an instantaneous torque on the portion of the object in contact with the surface  This torque does NO work as it is only in contact with the SAME portion of the object for an infinitesimal time.

Vector Product a X b  a ab  b ab  a bsin   is the angle between a and b X X

axbaxb b a b a axbaxb Right Hand Rule

j k i right handed axis system

Using the vector product Torque can be written as  r  F r = position vector from an origin to the point of contact of the force F unit vector in direction of rotational axis is ˆ r  ˆ F = ˆ 

F r Choose an origin then draw the position vector

If F is the total force then  is the total torque

Properties of Vector Product a X b  absin  a  a X b & b  ab aa  0 ab  ba d dt ab   da b  a db X X X X X XX

Angular Momentum L  r  p L depends on the choice of origin L  r  p  rp t L   rmv t  rmr  mr 2  I   L  I   ˆ  r  ptpt 

Forces acting on many particles that are rigidly fixed with respect to each other or an extended rigid body

Thus if only internal forces act  tot  0  int In general for any system made up of many objects that are fixed with respect to each other or an extended rigid object  tot  int  ext 

The vector product and the choice of origin defines the axis of rotation of the rigid body

Conservation of Total Angular Momentum If a system is isolated  tot  0  dL dt Conservation of Total Angular Momentum L tot,initial  L tot,final e.g. I 1i  1i  ˆ  I 2i  2i  ˆ  I 1f  1f  ˆ  I 2f  2f  ˆ 