Dynamics READ the Textbook Rotational Motion Part III

Dynamics READ the Textbook! Rotational Motion Part III – “What we Angular velocity know is a drop, what we don't know is an ocean. ” Angular “suvat” Kinetic energy – rotation Moments of Inertia KE – linear, rotation http: //www. hep. manchester. ac. uk/u/parkes/Chris_Parkes/Teaching. html October 2012 Chris Parkes

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

Rotational Motion This week – deal with components, next week with vectors Linear Rotational Position, x Velocity, v Acceleration Angle, θ angular velocity, ω angular acceleration, α Rotational “suvat” derive: constant angular acceleration

Rotational Kinetic Energy X-section of a rigid body rotating about an axis through O which is perpendicular to the screen ω m 1 m 2 r 1 O r 3 m 3 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 masses m quantities distance from rotation axis r – 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. frisbee solid sphere hula-hoop pc hard disk R neutron star R space station R 1 R 2

Moment of Inertia Calculations Systems of discrete particles 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 ? y 2 kg x 2 kg 1 m 2 m I = 2 x (2 x 12) + (2 x 22) = 12 kg m 2

Parallel-Axis Theorem • Moment of Inertia around an axis parallel to first at distance d – Co-ordinate system origin at centre-of-mass y. A • Proof x. A Since centre-of-mass is origin of co-ordinate system

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 R M M m h h m 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.

KEY POINTS • Assume flywheel is a solid cylinder R • Note – cable is light – hence ignore its mass M • 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 h m 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.

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=0 ω=0 h v No slipping so no work done against friction – Mechanical energy conserved