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Short Version : 10. Rotational Motion Polar coord ( r, ) 10.1. Angular Velocity & Acceleration θ̂ r̂ Average angular velocity ω ˆ ω ˆ ω t Right hand rule = angular displacement ( positive if CCW ) ω̂ // rotational axis in radians 1 rad = 360 / 2 = 57.3 (Instantaneous) angular velocity d ˆ ˆ ω ˆ ω lim ω ω t 0 t dt Angular speed: Circular motion: Linear speed: s r v ω̂ d dt ds r dt in radians 1ds v r dt r ω̂ Angular Acceleration We shall restrict ourselves to rotations about a fixed axis. (Instantaneous) angular acceleration at d d2 lim t 0 t dt d t2 ˆ α ω Trajectory of point on rotating rigid body is a circle, v / / θˆ a ar i.e. r = const. Its velocity v is always tangential: v r θˆ Its acceleration is in the plane of rotation ( ) : ω̂ d ˆ d θˆ dv r θ r a at a r d t d t dt d ˆ Tangential component: at r θ r θˆ dt Radial component: d θˆ v2 2 ar r rˆ r rˆ dt r d θˆ d rˆ dt dt rˆ Angular vs Linear Table 10.1 Angular & Linear Position, Velocity, & Acceleration Linear Quantity Angular position Position x Velocity Acceleration Angular Quantity v dx dt d v d2 x a dt d t2 Eqs for Constant Linear Acceleration 1 v0 v 2 v v0 a t v Angular velocity Angular Acceleration d dt d d dt d t2 Eqs for Constant Angular Acceleration 1 0 2 0 t 1 x x0 v0 t a t 2 2 0 0 t t 2 1 2 v 2 v02 2a x x0 2 02 2 0 10.2. Torque Rotational analog of force Torque : τ τˆ r F sin τ̂ plane of r & F r to F right-hand rule. [ ] = N-m ( not J ) r sin 10.3. Rotational Inertia & the Analog of Newton’s Law Linear acceleration: F ma Rotating baton (massless rod of length R + ball of mass m at 1 end): R Ft Tangential force on ball: Ft m at m R m R2 I I mR 2 = moment of inertia = rotational inertia of the baton Calculating the Rotational Inertia Rotational inertia of discrete masses I m i ri 2 i ri = perpendicular distance of mass i to the rotational axis. Rotational inertia of continuous matter I r 2 d m r 2 r dV r = perpendicular distance of point r to the rotational axis. ( r) = density at point r. Example 10.4. Dumbbell A dumbbell consists of 2 equal masses m = 0.64 kg on the ends of a massless rod of length L = 85 cm. Calculate its rotational inertia about an axis ¼ of the way from one end & perpendicular to it. GOT IT? 10.2 Would I (a) increase L 2 3 L 2 5 I m m L2 8 4 4 5 2 0.64 kg 0.85 m 0.29 kg m2 8 (b) decrease (c) stay the same if the rotational axis were (b) (1) at the center of the rod (a) (2) at one end? Example 10.5. Rod Find the rotational inertia of a uniform, narrow rod of mass M and length L about an axis through its center & perpendicular to it. I r 2 d m r 2 dV M 1 3 x L 3 L /2 L /2 M 2 L 12 L /2 x 2 dx L /2 L /2 L /2 x2 M dx L Example 10.6. Ring Find the rotational inertia of a thin ring of radius R and mass M about the ring’s axis. 2 R2 R d I R2 d m M R2 2 0 2 0 d M 2 R R2 d m M R2 Pipe of radius R & length L : I L 0 R3 2 0 R R d d z 2 M 2 R L M 2 L M R 2 R 2 d m 2 R L I = MR2 for any thin ring / pipe Example 10.7. Disk Find the rotational inertia of a uniform disk of radius R & mass M about an axis through its center & perpendicular to it. I r2 d m M R2 d m r d d r I 2 0 d R 0 3 r dr 1 2 R 4 4 1 M R2 2 R 0 r 2 dM dM 2 r d r Table 10.2. Rotational Inertia Parallel - Axis Theorem Parallel - Axis Theorem: I I cm M d 2 Ex. Prove the theorem for a set of particles. Example 10.9. Into the Well A solid cylinder of mass M & radius R is mounted on a frictionless horizontal axle over a well. A rope of negligible mass is wrapped around the cylinder & supports a bucket of mass m. Find the bucket’s acceleration as it falls into the well. Let downward direction be positive. Bucket: Cylinder: Fnet mg T m a a T R I I R mg I a ma 2 R g a 1 I mR 2 g M 1 2m T I a R2 GOT IT? 10.4. Two masses m is connected by a string that passes over a frictionless pulley of mass M. One mass rests on a frictionless table; the other vertically. Is the magnitude of the tension force in the vertical section of the string (a) greater than, (b) equal to, or (c) less than in the horizontal? Explain. (a): There must be a net torque to increase the pulley’s clockwise angular velocity. 10.4. Rotational Energy Rotational kinetic energy = sum of kinetic energies of all mass elements, taken w.r.t the rotational axis. Set of particles: K 1 1 2 m v mi ri 2 2 i i 2 i 2 i dK 1 I 2 2 1 1 2 2 dm v dm r 2 2 K rot dK 1 1 2 r d m 2 2 2 K rot 1 I 2 2 r2 d m Energy & Work in Rotational Motion Work-energy theorem for rotations: 2 W d K rot 1 1 1 I 2f I i2 2 2 10.5. Rolling Motion V = velocity of CM. ui = velocity relative to CM. Composite object: Ktotal 1 mi v i2 2 i 1 2 m V u i i 2 i 1 1 M V 2 mi ui2 2 2 i 1 mi V 2 ui2 2V ui 2 i mi ui i d mi ri R cm 0 dt i Ktotal Kcm Kinternal Moving wheel: K total 1 1 M V 2 u 2 dm 2 2 Ktotal 1 1 M V 2 2 r 2 dm 2 2 1 1 M V 2 I cm 2 2 2 is w.r.t. axis thru cm 1 1 M V 2 I cm 2 2 2 Moving wheel: Ktotal Rolling wheel: X R V R V = velocity of CM is w.r.t. axis thru CM Example 10.12. Rolling Downhill A solid ball of mass M and radius R starts from rest & rolls down a hill. Its center of mass drops a total distance h. Find the ball’s speed at the bottom of the hill. Initially:E0 Ktrans 0 K rot 0 U 0 M g h Finally: E Ktrans K rot U 1 1 M v2 I 2 2 2 1 12 v M v2 M R2 2 25 R E E0 v 10 gh 7 2 7 M v2 10 2g h sliding ball Note: v is independent of M & R