Rotational Kinetic Energy and Inertia

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

Recall: Rotational Kinematics And for a point at a distance R from the rotation axis: • xT = RvT= R aT = R Angular Linear

Comment on Axes and Signs(i.e. what is positive and negative) • Whenever we talk about rotation, it is implied that there is a rotation “axis”. • This is usually called the “z” axis (we usually omit the z subscript for simplicity). • Counter-clockwise (increasing q) is usuallycalled positive. • Clockwise (decreasing q) is usuallycalled negative. +w 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 w. • Mass has speed v = w R • Mass has kinetic energy • K = ½ M v2 • K = ½ M w2 r2 • 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 w. • Each “piece” of disk has speed vi=wri • Each “piece” has kinetic energy • Ki = ½ mi v2 • = ½ miw2 ri2 • Combine all the pieces • SKi = S ½ miw2 r2 • = ½ (S mi ri2) w2 • = ½ I w2 ri I is called the Rotational Inertia

Rotational Inertia I • Tells how much “work” is required to get object spinning. Just like mass tells you how much “work” is required to get object moving. • Ktran = ½ m v2 Linear Motion • Krot = ½ I w2Rotational Motion • I S miri2(units kg-m2) • Note! Rotational Inertia depends on what axis you are spinning about (the ri in the equation).

Rotational Inertia Table • For objects with finite number of masses, use I = S m r2. • For “continuous” objects, use table below:

Summary • Rotational Kinetic Energy Krot = ½ I w2 • Rotational Inertia I = S miri2 • Energy is Still Conserved!

Pulley Example • 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? • We will use conservation of energy: • Remember: • Ktrans = ½ m v2 • Krot= ½ I 2 • U = m g y 5kg 8 kg 1.5 m

Pulley Example • 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? • Initial Krot and Ktran is zero. • Final U is zero. 5kg 8 kg 1.5 m

Pulley Example • 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? • Remember: • I = ½ m R2 for a disk • v = R so  = v/R 5kg 8 kg 1.5 m

Pulley Example • 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? • R2 cancels from the last term. • Simplify: 5kg 8 kg 1.5 m

Pulley Example • 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? • Solve: 5kg vf = 4.73 m/s 8 kg 1.5 m

Torque • Rotational effect of force. Tells how effective force is at twisting or rotating an object. • t = r F sin q=  r Fperpendicular • Units: N-m • Sign: CCW rotation is positive F  r

Work Done by Torque • Recall W = F d cos q • For a wheel: • W = Ftangential s = Ftangential r q (q in radians) W = t q • P = W/t = t q/t P = t w

Equilibrium • Conditions for Equilibrium: • S F = 0 Translational Equilibrium (Center of Mass) • St = 0 Rotational Equilibrium • May choose any axis of rotation…. But Choose Wisely!

Equilibrium Example • 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? • First, draw a FBD: F2 Fg F1

Equilibrium Example • 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? • Write down F = 0: -F1 + F2 – Fg = 0 -F1 + F2 – (50 kg)(9.8 m/s2) = 0 Note: there are two unknowns so we need another equation…

Equilibrium Example • 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? • Pick a pivot point: F2 Fg F1

Equilibrium Example • 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? • Write down  = 0: F1 (0 m) + F2 (1.2 m) – (50 kg)(9.8 m/s2)(4.6 m) = 0 F2 (1.2 m) – (50 kg)(9.8 m/s2)(4.6 m) = 0

Equilibrium Example • 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? • Solve: -F1 + F2 – (50 kg)(9.8 m/s2) = 0 -F1 + (1878 N) – (50 kg)(9.8 m/s2) = 0 F1 = 1388 N F2 (1.2 m) – (50 kg)(9.8 m/s2)(4.6 m) = 0 F2 = 1878 N

Summary • Torque is a Force that causes rotation • t = F r sin q • Work done by torque: W = t q • Equilibrium • S F = 0 • S t = 0 • May choose any axis.

Rotational Kinetic Energy and Inertia