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

2. Rotational Form Newton’s 2 nd Law  = I  è Torque is an amount of twist provided by a force »Sign: positive = CCW è Moment of Inertia is like mass » Large I means it is hard to start or stop from spinning l Problems Solved Like F=ma è Draw FBD  Write  = I 

3. 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? è Draw FBD for each object: m3m3 m1m1 m2m2 m3m3 m1m1 m2m2 FNFN FgFg T1T1 T1T1 T2T2 T2T2 FgFg

4. 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? è Newton’s Second Law for each object: m3m3 m1m1 m2m2 Mass 1:  F = ma T 1 = m 1 a Mass 2:  = I  T 2 R – T 1 R = I  Mass 3:  F = ma m 3 g – T 2 = m 3 a note: we do not need the vertical direction for m 1

5. 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? è Solve equations 1 and 3 for T 1 and T 2. è Substitute into equation 2. è Replace  with a/R and I with ½ m 2 R 2. è Simplify: m3m3 m1m1 m2m2 (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

6. 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? è Solve for a: m3m3 m1m1 m2m2 = 2.8 m/s 2

7. Rolling (“without slipping”) l Static friction f causes rolling “without slipping”. l Friction causes object to roll, but if it rolls without slipping friction does NO work!  W = F d cos  but d is zero for point in contact è No 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:  x =  R  v =  R  a =  R  R I M

8. 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. è Use Conservation of Energy: H

9. 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. è Recall: »I = ½ M R 2 »  = v / R H

10. 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. è Solve for v: H = 5.11 m/s

11. Summary  = I  è Newton’s Second Law for Rotations è Torque, Moment of Inertia, FBDs l Rolling è Energy is Conserved è Need to include Translational and Rotational Energy