Lecture 13: Rotational Kinetic Energy and Rotational Inertia l Review: Rotational Kinematics l Rotational Kinetic Energy l Rotational Inertia l Torque l Rotational Dynamics l Equilibrium

Recall: Rotational Kinematics Angular Linear And for a point at a distance R from the rotation axis: x T =  R  v T =  R  a T =  R

Comment on Axes and Signs (i.e. what is positive and negative) l Whenever we talk about rotation, it is implied that there is a rotation “axis”. l This is usually called the “z” axis (we usually omit the z subscript for simplicity). Counter-clockwise (increasing  ) is usually called positive. Clockwise (decreasing  ) is usually called negative. z 

Rotational Kinetic Energy Consider a mass M on the end of a string being spun around in a circle with radius R and angular frequency   Mass has speed v =  R è Mass has kinetic energy »K = ½ M v 2 »K = ½ M  2 r 2 l Rotational Kinetic Energy is energy due to circular motion of object. M

Kinetic Energy of Rotating Disk Consider a disk with radius R and mass M, spinning with angular frequency   Each “piece” of disk has speed v i =  r i è Each “piece” has kinetic energy »K i = ½ m i v 2 » = ½ m i  2 r i 2 è Combine all the pieces »  K i =  ½ m i  2 r 2 » = ½ (  m i r i 2 )  2 » = ½ I  2 riri I is called the Rotational Inertia

Rotational Inertia I l Tells how much “work” is required to get object spinning. Just like mass tells you how much “work” is required to get object moving. è K tran = ½ m v 2 Linear Motion  K rot = ½ I  2 Rotational Motion I   m i r i 2 (units kg-m 2 ) l Note! Rotational Inertia depends on what axis you are spinning about (the r i in the equation).

Rotational Inertia Table For objects with finite number of masses, use I =  m r 2. l For “continuous” objects, use table below:

Summary Rotational Kinetic Energy K rot = ½ I  2 Rotational Inertia I =  m i r i 2 l Energy is Still Conserved!

Pulley Example l A solid disk wheel is fixed at its center and has a string wrapped around it and then attached to a block. The mass of the pulley is 5 kg, the radius of the pulley is 0.2 m, and the mass of the block is 8 kg. After the block has fallen 1.5 m, how fast is the block moving and how fast is the disk rotating? è We will use conservation of energy: è Remember: »K trans = ½ m v 2 »K rot = ½ I  2 »U = m g y 5 kg 8 kg 1.5 m

Pulley Example l A solid disk wheel is fixed at its center and has a string wrapped around it and then attached to a block. The mass of the pulley is 5 kg, the radius of the pulley is 0.2 m, and the mass of the block is 8 kg. After the block has fallen 1.5 m, how fast is the block moving and how fast is the disk rotating? è Initial K rot and K tran is zero. è Final U is zero. 5 kg 8 kg 1.5 m

Pulley Example l A solid disk wheel is fixed at its center and has a string wrapped around it and then attached to a block. The mass of the pulley is 5 kg, the radius of the pulley is 0.2 m, and the mass of the block is 8 kg. After the block has fallen 1.5 m, how fast is the block moving and how fast is the disk rotating? è Remember: »I = ½ m R 2 for a disk »v =  R so  = v/R 5 kg 8 kg 1.5 m

Pulley Example l A solid disk wheel is fixed at its center and has a string wrapped around it and then attached to a block. The mass of the pulley is 5 kg, the radius of the pulley is 0.2 m, and the mass of the block is 8 kg. After the block has fallen 1.5 m, how fast is the block moving and how fast is the disk rotating? è R 2 cancels from the last term. è Simplify: 5 kg 8 kg 1.5 m

Pulley Example l A solid disk wheel is fixed at its center and has a string wrapped around it and then attached to a block. The mass of the pulley is 5 kg, the radius of the pulley is 0.2 m, and the mass of the block is 8 kg. After the block has fallen 1.5 m, how fast is the block moving and how fast is the disk rotating? è Solve: 5 kg 8 kg 1.5 m v f = 4.73 m/s

Torque l Rotational effect of force. Tells how effective force is at twisting or rotating an object.  = r F sin  r F perpendicular è Units: N-m è Sign: CCW rotation is positive F  r

Work Done by Torque Recall W = F d cos  l For a wheel: è W = F tangential s = F tangential r  (  in radians) W =    P = W/t =  /t  P 

Equilibrium l Conditions for Equilibrium:   F = 0 Translational Equilibrium (Center of Mass)    = 0 Rotational Equilibrium »May choose any axis of rotation…. But Choose Wisely!

Equilibrium Example l A 50 kg diver stands at the end of a 4.6 m diving board. Neglecting the weight of the board, what is the force each pivot exerts on the diving board? è First, draw a FBD: F1F1 F2F2 FgFg

Equilibrium Example l A 50 kg diver stands at the end of a 4.6 m diving board. Neglecting the weight of the board, what is the force each pivot exerts on the diving board? è Write down  F = 0: -F 1 + F 2 – F g = 0 -F 1 + F 2 – (50 kg)(9.8 m/s 2 ) = 0 Note: there are two unknowns so we need another equation…

Equilibrium Example l A 50 kg diver stands at the end of a 4.6 m diving board. Neglecting the weight of the board, what is the force each pivot exerts on the diving board? è Pick a pivot point: F1F1 F2F2 FgFg

Equilibrium Example l A 50 kg diver stands at the end of a 4.6 m diving board. Neglecting the weight of the board, what is the force each pivot exerts on the diving board? è Write down  = 0: F 1 (0 m) + F 2 (1.2 m) – (50 kg)(9.8 m/s 2 )(4.6 m) = 0 F 2 (1.2 m) – (50 kg)(9.8 m/s 2 )(4.6 m) = 0

Equilibrium Example l A 50 kg diver stands at the end of a 4.6 m diving board. Neglecting the weight of the board, what is the force each pivot exerts on the diving board? è Solve: F 2 (1.2 m) – (50 kg)(9.8 m/s 2 )(4.6 m) = 0 F 2 = 1878 N -F 1 + F 2 – (50 kg)(9.8 m/s 2 ) = 0 -F 1 + (1878 N) – (50 kg)(9.8 m/s 2 ) = 0 F 1 = 1388 N

Summary l Torque is a Force that causes rotation   = F r sin   Work done by torque: W =  l Equilibrium   F = 0   = 0 »May choose any axis.