Dynamics Chris Parkes October 2012 Dynamics Rotational Motion Angular velocity Angular “suvat” Kinetic energy – rotation Moments of Inertia KE – linear, rotation Part III – “What we know is a drop, what we don't know is an ocean.” READ the Textbook!

Rotation of rigid body about an axis θ in radians –2πr is circumference Angular velocity –same for all points in body Angular acceleration θ s = rθ r v r is perpendicular distance to axis dθ r ds

Rotational Motion LinearRotational Position, xAngle, θ Velocity, vangular velocity, ω Accelerationangular acceleration, α This week – deal with components, next week with vectors Rotational “suvat” derive: constant angular acceleration

Rotational Kinetic Energy O m1m1 r1r1 m2m2 m3m3 r3r3 r2r2 ω X-section of a rigid body rotating about an axis through O which is perpendicular to the screen r 1 is perpendicular dist of m 1 from axis of rotation r 2 is perpendicular dist of m 2 from axis of rotation r 3 is perpendicular dist of m 3 from axis of rotation

Moment of Inertia, I corresponding angular quantities for linear quantities –x  ; v  ; p  L –Mass also has an equivalent: moment of Inertia, I –Linear K.E.: –Rotating body v , m  I: –Or p=mv becomes: Conservation of ang. mom.: e.g. frisbeesolid sphere hula-hoop pc hard diskneutron star space station R  R   R1R1 R2R2 masses m distance from rotation axis r

Four equal point masses, each of mass 2 kg are arranged in the xy plane as shown. They are connected by light sticks to form a rigid body. What is the moment of inertia of the system about the y-axis ? Moment of Inertia Calculations Systems of discrete particles 2 kg 1m 2m I = 2 x (2 x1 2 ) + (2 x 2 2 ) = 12 kg m 2 y x

Parallel-Axis Theorem xAxA yAyA Proof Since centre-of-mass is origin of co-ordinate system Moment of Inertia around an axis axis parallel to first at distance d –Co-ordinate system origin at centre-of-mass

Table 9.2 Y& F p 291 (14 th Edition)

Translation & Rotation Combining this lecture and previous ones Break problem into –Velocity of centre of mass –Rotation about axis ω

Rolling without Slipping Bicycle wheel along road Centre – pure translation Rim –more comlex path known as cycloid If no slipping

R M m h R M m h  v A light flexible cable is wound around a flywheel of mass M and radius R. The flywheel rotates with negligible friction about a stationary horizontal axis. An object of mass m is tied to the free end of the cable. The object is released from rest at a distance h above the floor. As the object falls the cable unwinds without slipping. Find the speed of the falling object and the angular speed of the flywheel just as the object strikes the floor.

R M m h  v KEY POINTS Assume flywheel is a solid cylinder Note – cable is light – hence ignore its mass Note – cable unwinds without slipping - speed of point on rim of flywheel is same as that of the cable If no slipping occurs, there must be friction between the cable and the flywheel But friction does no work because there is no movement between the cable and the flywheel

Example A bowling ball of radius R and mass M is rolling without slipping on a horizontal surface at a velocity v. It then rolls without slipping up a slope to a height h before momentarily stopping and then rolling back down. Find h. v h v = 0 ω = 0 No slipping so no work done against friction – Mechanical energy conserved