Summary Lecture 12 Rotational Motion 10.8Torque 10.9Newton 2 for rotation Work and Power 11.2Rolling motion Rotational Motion 10.8Torque 10.9Newton 2 for rotation Work and Power 11.2Rolling motion Problems: Chap. 10: 21, 28, 33, 39, 49, 54, 67 Problems: Chap. 10: 21, 28, 33, 39, 49, 54, 67 Today 20-minute test on material in lectures 1-7 at end of lecture Today 20-minute test on material in lectures 1-7 at end of lecture

Some Rotational Inertia

PLUS Axis of Rotation h Parallel-axis Theorem CM The rotational inertia of a body about any parallel axis, is equal to its R.I. about an axis through its CM, R.I. of its CM about a parallel axis through the point of rotation I = I CM + Mh 2

One rotation about yellow axis involves one rotation of CM about this axis plus one rotation of body about CM. I = I cm + Mh 2 Proof of Parallel-axis Theorem h

RI of CM about suspension point, distance R away is MR 2. So total RI is 2MR 2 Example What is it about here? R RI of ring of mass M about CM is MR 2

The Story so far... Rotational Variables , ,  relation to linear variables vector nature Rotational kinematics with const.  Rotation and Kinetic Energy Analogue equations to linear motion Rotational Inertia

Torque …is the “turning ability” of a force Where would you put the door knob? here?or here? F F r r The magnitude of the torque is Fr, and this is greater here!

If F and r are perpendicular  =  r  F  (Unit: N m) Torque Axis r F The same F at larger r has bigger turning effect.

F  r Torque is a vector  = r F sin   =  r  x perp component of F Direction of  : Perpendicular to r and F Sense: Right-hand screw rule (out of screen) (Hint: Which way would it accelerate the body? Sense is same as change in ang vel. .)  = r x F In General Fsin 

Newton 2 for Rotational motion For Translational motion we had: For Rotational motion we expect:

What is I for rotating object? Example  = I  (Newton 2)  I =  /   I = 6400/1.2  I = 5334 kg m 2  Observe ang accel of 1.2 rad s -2 Apply a force of 3200 N F r Distance r = 2 m from axis   = r x F = 6400 N m

R Work done by a torque  which rotates a body through an angle   w =   cf If  is constant w =  Power P =  Work    cf P = F.v

Example A car engine has a power of 100 HP at 1800 RPM What is the torque provided by the engine? 1 HP = 746 W  P = W P = .    = P/    = 74600/ 188   = 397 N m

v cm t = 2  R But in turning one revolution (2  radian) in time t,  = 2  /t So that t = 2  /  2R2R v cm 2  /  = 2  R v cm =  R Rolling Motion R

total energy (KE) = ½ mv 2 + ½ v 2  mR 2 /R 2 Object of radius R The kinetic energy is shared between translational and rotational motion. = ½ mv 2 (1 +  ) Cons. Energy says PE initial = KE final mgh = K trans + K rot = ½ mv cm 2 + ½ I  2 h remember v cm = R  or  = v cm /R so mgh = ½ mv 2 + ½ Iv 2 /R 2  is the coefficient in the expression for the Rotational Inertia I I =  mR 2  is the coefficient in the expression for the Rotational Inertia I I =  mR 2 = ½ mv 2 + ½ mv 2 

h The kinetic energy is shared between translational and rotational motion. mgh = K total = K trans + K Rot = ½ mv cm 2 + ½ mv cm 2  The fraction of KE that is translational is The larger , the more of the available energy goes into rotational energy, and the smaller the centre of mass velocity h = ½ mv cm 2 (1 +  )