Consider a wheel that rolls without slipping. The centre of the wheel moves with pure translation while a point on a rim will trace out a cycloid.
A rolling wheel is a combination of pure rotation and pure translational motion. v=2vcm v=vcm v=vcm vcm = + vcm v=vcm v=-vcm v=0
A person watching the wheel sees the com move with constant velocity vcm. The wheel rotates through an angle θ while moving a distance s (given by the arc length). P' vcm vcm s P s
So that, P' vcm vcm s P s
So that, • Equivalently for the top of the wheel, P' v=2vcm v=2vcm vcm vcm s P s
Kinetic Energy of the wheel • The KE wrt a stationary is, • From the parallel-axis theorem, Rotation about the point P P' v=2vcm v=2vcm vcm vcm s P s
Example • A roller (solid cylindrical disk) of mass M=50kg and has a radius R=1.5m rolls across a smooth pitch at a speed of 15m/s. (a) What is the speed at the top of the roller, (b) its angular speed, (c) the total kinetic energy?
(a) The speed of a rolling object is the velocity of the com. • (b) • (c)
The angular momentum is analogous to the linear momentum. • The angular momentum is defined as,
Newton’s 2nd law can be written in angular form. • Given, • Then, • That is, the vector sum of the torque is equal to the rate of change of angular momentum. (single particle) (single particle)
For a system of particles the angular momentum of the system is, • Thus, (system of particles)
For a system of particles the angular momentum of the system is, • Thus, • For a rigid body, (system of particles)
The conservation momentum also holds for angular momentum. • If no net external torque acts on a system then,
Conservation of Angular momentum Angular momentum is conserved. The momentum of inertia decreases when the skater pulls his arms inward. As a result the angular momentum must increase
Example • The figure shows a 0.1kg mass attached to a string which passes through a vertical tube. The mass initial rotated in the horizontal plane (a circle of radius 0.25m) at an angular speed of 5rad/s. The string is pulled downward so that radius of the string is now 0.15m. Calculate the new angular speed.
Sol: r r ωi ωf