Chap. 10: Rotational Motion

Chap. 10: Rotational Motion I. Rotational Kinematics II. Rotational Dynamics - Newton s Law for Rotation III. Angular Momentum Conservation (Chap. 10) 1

Newton s Laws for Rotation n e t I 3 rd part [N m] Chap. 10 nd part [kg m ] 1 st part [s ] K rot = (1/) I w (K = (1/) m v ) Angular Momentum Conservation

a x Rotational Dynamics [torque and moment of inertia] (b) m I F (a) F Speeding up l a l b a x F F cos cos m Mass (kg) is a measure of the inertia of a body. a x F F l ; l I (torque ) n e t F l I F l (magnitude ) 3

Vector Nature of Angular Quantities Kinematical variables to describe the rotational motion: Angular position, velocity and acceleration c.w. or c.c.w. rotation (like +x or x direction in 1D) z Vector natures! w kˆ kˆ l R (rad) d kˆ dt dw kˆ dt (rad/s ) (rad/s ) x R.-H. Rule y 4

Speeding Up Acceleration vector z Velocity vector R.-H. Rule y x 5

Slowing Down Acceleration vector z Velocity vector R.-H. Rule y x 6

net ( r F ) I Magnitude (b) Speeding up F r l b Direction I Lever arm : l (Perpendicular or r distance...) 7

= r x F x F y z F 10

Cross Product: General 11

A grindstone in the shape of a solid disk (mass M = 50.0 kg and radius R = 0.60 m) is rotating at 850 rev/min. You press an ax against the rim with a normal force of 160 N and the grindstone comes to rest in 7.50 s. Find the coefficient of friction between the ax and the grindstone. ISEE Identifying Find: Ff k FP Need to know F f What kind of motion? Rigid body rotation Kinematic eqs. & = /I ISEE Setting up ISEE Solve the equations ISEE 1

Setting up Vector Nature of w z R.-H. Rule w 0 = 850 rev/min Slowing down F P F f = 90 o r I x y Rigid body rotation Slowing down by friction Direction of friction? Direction of w? [out of page] Direction of? [into page] Torque due to friction magnitude &direction 13

Solutions Rigid body rotation Slowing down by friction Direction of friction? Direction of w? [out of page] Direction of? [into page] Torque due to friction magnitude &direction w 0 = 850 rev/min Slowing down F P F f = 90 o r F I w w 0 t r I k F F f P R F P I RF P MR /w t RF P 0. 0 kg0. 60 m850 rev min rad s 50 60 rev min 7. 50 s160 N 0. 48. 14

Note: sign of and S 1 (5 0.0 N )(0.3 0 0 F 1 F ( R 1 (50.0 ( R s in 90 sin 60 ) N )(0.500 m) 1 5.0 N ) m)(0.866) m r 1 r net 1.7 [ 1 1 N m ] [ ] [ ] [ ] [(15.0 N m) (1.7 N m)][ ] 6.7 N m[ ] 6.7 N m [ ] 18

Note: = F R sin Le v e r a rm (Pe rpe ndic : l ula r o r R dis ta nce...) R F ( R s i n ) F R R ( F s i n ) R F R 1

Torque due to Gravity? r F? We assume: Center of mass (CM or cm) = Center of gravity (if uniform gravity)

Torque due to Gravity? r F? mass m l 4

Net Torque due to Gravity? One extra sphere! Do we need a new concept? No! mass m l n e t r o d s p h e r e Solid sphere (M, r) 6

Rolling Motion (w/o slipping) PHYSICS, R. Serway and R. Beichner, 5 th ed. 8

Rolling Motion w/o Slipping v = R w w w : common for all points v cm v cm = R w P is: (a) the point of contact with the ground; (b) instantaneously rest relative to the ground. a pure rotation about the instantaneous axis through P at the instant. 9

Newton s nd Law for Rolling Motion w/o Slipping F n e t m a n e t I K trans + K Rotational rot = K Motion rolling 30

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Example 3 What will be the speed of each of the following objects when it reaches the bottom of an incline if it starts from rest at a vertical height H and rolls without slipping? Find the speed using the conservation of mechanical energy. (i) Solid sphere (M, R 0 ) (ii) Solid cylinder (M, R 0, l) (iii) Thin hoop (M, R 0, l) 3

Example 3 (cont d) K i U where U MgH i i K 1 f 0, U Mv U f f 1 0, I K cm f w K trans K rot (i) Solid sphere (M, R 0 ) (ii) Solid cylinder (M, R 0, l) (iii) Thin hoop (M, R 0, l) Rotational Inertia: I cm Rotational Inertia: I cm 33

Example 10.7 Example 4 Analyze the rolling sphere in terms of forces and torques: find the magnitudes of the velocity v and the friction force F fr. 1. F.B.D.? Rolling Motion (w/o slipping). Is F fr static or kinetic friction? Static Friction No work No energy loss K+U=constant P is: (a) the point of contact with the ground; (b) instantaneously rest relative to the ground. Thus: The motion of the wheel is a pure rotation about the instantaneous axis through P at the instant. 34

Examples in Textbook Analyze the rolling sphere in terms of forces and torques: find the magnitudes of the velocity v and. 36

Applications of Newton s Laws for Rotation Equilibrium Analysis F net net 0 0 Non-Equilibrium Analysis F net net ma I 37

More Problems 38

Example 1 Calculate the torque on the.00-m long beam due to a 50.0 N force (top) about (a) point C (= c.m.) (b) point P Calculate the torque on the.00-m long beam due to a 60.0 N force about (a) point C (= c.m.) (b) point P Calculate the torque on the.00-m long beam due to a 50.0 N force (bottom) about (a) point C (= c.m.) (b) point P 39

Example 1 (cont d) Calculate the net torque on the.00-m long beam about (a) point C (= c.m.) (b) point P 40

Problem Determine the net torque (magnitude and direction) on the.00-m-long beam about: a. (15 pts) point C (the CM position); b. (10 pts) point P at one end. i net r i i Fi i? 50.0 N = 70.0 o 60.0 N = 50.0 o C P = 33.0 o 50.0 N 41

Problem 3 Determine the net torque (magnitude and direction) on the disk about pin. i net r i i Fi i? F 1 50 N 1 =40 o =70 o F 5 N 4

BACKUP 43

Non-Equilibrium Analysis Solid disk (M, R 0 ) m 47

M g H 1 M v 1 I cm w I cm = Large v = Small 48

Example 6 - Challenge A thin uniform stick of mass M and length l is positioned vertically, with its tip on a frictionless table. It is released and allowed to slip and fall. See the figure. Determine the speed of its center of mass just before it hits the table. (i) The CM must fall straight down, because there is no friction. (ii) No vertical motion. 49

Example 6 - Challenge There is no friction. The CM must fall straight down. = + K i U i K f U f, whe re K i 0, U f 0, K f K tra ns K ro t Mg l 1 Mv cm 1 I cm w cm 50

Example Two blocks are connected by a light string passing over a pulley of radius 0.50 m and mass.00 kg. The blocks move to the right with a constant acceleration on inclines with frictionless surface. (a) Draw the free-body diagram for each of the two blocks; (b) Draw the free-body diagram for the pulley; (c) Determine the tensions in the two parts of the string and acceleration of the blocks. 51

Practice Problem 4 A uniform disk turns at 7.00 rev/s around a frictionless spindle. A nonrotating rod, of the same mass (m) as the disk and length (l) equal to the disk s diameter, is dropped onto the d i s k. T h e y t h e n b o t h turnaround the spindle with their centers superposed. There is no slipping between the rod and the disk. What is the mo ment of inertia of the disk+rod system about the axis? 5

K.E. of Rolling Motion w/o Slipping A rolling wheel A rolling solid cylinder where I cm = (1/) MR w I rolling = I cm + MR = (3/) MR K rolling = (1/) I rolling w = (1/) I cm w + (1/) M (R w ) w = K rot + K trans Faster v cm Instantaneously rest Instantaneous axis K rolling = K rot + K trans 53

Speed of Yo-Yo A sting is wrapped around a uniform solid cylinder of mass M (= 0.500 kg) and radius R (= 0.00 m), and the cylinder starts falling from rest. Find the speed after the cylinder has descended h = 0.800 m. Use mechanical energy conservation, 54

Speed of Yo-Yo A sting is wrapped around a uniform solid cylinder of mass M (= 0.500 kg) and radius R (= 0.00 m), and the cylinder starts falling from rest. (a) Draw the free-body diagram for the cylinder while it descends. (b) Find its acceleration and the tension in the string. (c) Find the speed after the cylinder has descended h = 0.800 m. 55

Speed of Yo-Yo F.B.D. 56

Yo-Yo Challenge A sting is wrapped around a hollow cylinder of mass M (= 0.500 kg), inner radius R 1 (= 0.100 m), outer radius R (= 0.00 m). The cylinder starts falling from rest. (a) Draw the free-body diagram for the cylinder while it descends. Also find: (b) its acceleration; (c) the tension in the string; (d) the speed after the cylinder has descended h = 0.800 m. 57