Lecture 14 Rolling Objects l Rotational Dynamics l

Lecture 14: Rolling Objects l Rotational Dynamics l Rolling Objects and Conservation of Energy l Examples & Problem Solving

Rotational Form Newton’s l nd 2 Law =I èTorque is an amount of twist provided by a force » Sign: positive = CCW èMoment of Inertia is like mass » Large I means it is hard to start or stop from spinning l Problems Solved Like F=ma èDraw FBD èWrite = I

Blocks & Pulley Example l In the system shown to the right, m 1 = 3 kg, m 2 = 4 kg, m 3 = 2 kg, and the radius of the pulley is R = 0. 1 m. The pulley is a solid massive disk; the tabletop is frictionless. What is the acceleration of the falling block? m 2 m 3 èDraw FBD for each object: T 2 FN T 1 m 1 Fg m 2 T 2 m 3 Fg m 1

Blocks & Pulley Example l In the system shown to the right, m 1 = 3 kg, m 2 = 4 kg, m 3 = 2 kg, and the radius of the pulley is R = 0. 1 m. The pulley is a solid massive disk; the tabletop is frictionless. What is the acceleration of the falling block? m 2 m 3 èNewton’s Second Law for each object: Mass 2: Mass 1: Mass 3: F = ma = I F = ma T 1 = m 1 a T 2 R – T 1 R = I m 3 g – T 2 = m 3 a note: we do not need the vertical direction for m 1

Blocks & Pulley Example l In the system shown to the right, m 1 = 3 kg, m 2 = 4 kg, m 3 = 2 kg, and the radius of the pulley is R = 0. 1 m. The pulley is a solid massive disk; the tabletop is frictionless. What is the acceleration of the falling block? m 2 m 3 èSolve equations 1 and 3 for T 1 and T 2. èSubstitute into equation 2. èReplace with a/R and I with ½ m 2 R 2. èSimplify: (m 3 g – m 3 a)R – (m 1 a)R = (½ m 2 R 2) (a/R) m 3 g – m 3 a – m 1 a = ½ m 2 a m 3 g = (m 3 + m 1 + ½ m 2) a note: the radius cancels m 1

Blocks & Pulley Example l In the system shown to the right, m 1 = 3 kg, m 2 = 4 kg, m 3 = 2 kg, and the radius of the pulley is R = 0. 1 m. The pulley is a solid massive disk; the tabletop is frictionless. What is the acceleration of the falling block? m 2 m 1 m 3 èSolve for a: = 2. 8 m/s 2

Rolling (“without slipping”) l l Static friction f causes rolling “without slipping”. Friction causes object to roll, but if it rolls without slipping friction does NO work! èW = F d cos but d is zero for point in contact èNo dissipated work, energy is conserved l Consider CM motion and rotation about the CM separately when solving this problem. l Translational and Rotational Motion related: èx = R èv = R èa = R I M R

Energy Conservation Example l Consider a cylinder with radius R = 0. 5 m and mass M = 3 kg rolling down a ramp. Determine its speed after it drops a distance H = 2 m. èUse Conservation of Energy: H

Energy Conservation Example l Consider a cylinder with radius R = 0. 5 m and mass M = 3 kg rolling down a ramp. Determine its speed after it drops a distance H = 2 m. èRecall: » I = ½ M R 2 » =v/R H

Energy Conservation Example l Consider a cylinder with radius R = 0. 5 m and mass M = 3 kg rolling down a ramp. Determine its speed after it drops a distance H = 2 m. èSolve for v: H = 5. 11 m/s

Summary l =I èNewton’s Second Law for Rotations èTorque, Moment of Inertia, FBDs l Rolling èEnergy is Conserved èNeed to include Translational and Rotational Energy