Summary Lecture 11 Rotational Motion 10.5Relation between angular and linear variables 10.6Kinetic Energy of Rotation 10.7Rotational Inertia 10.8Torque 10.9Newton 2 for rotation Work and Power Rotational Motion 10.5Relation between angular and linear variables 10.6Kinetic Energy of Rotation 10.7Rotational Inertia 10.8Torque 10.9Newton 2 for rotation Work and Power Problems: Chap.10: 6, 7, 16, 21, 28, 33, 39, 49 Tomorrow 12 – 2 pm PPP “Extension” lecture. Room 211 podium level Turn up any time Tomorrow 12 – 2 pm PPP “Extension” lecture. Room 211 podium level Turn up any time This Friday 20-minute test on material in lectures 1-7 during lecture This Friday 20-minute test on material in lectures 1-7 during lecture

Relating Linear and Angular variables  r s =  r Need to relate the linear motion of a point in the rotating body with the angular variables  and s s

s =  r  v r  and v V, r, and  are all vectors. Although magnitude of v =  r. The true relation is v =  x r Not quite true.  s Relating Linear and Angular variables

v =  x r  r v Grab first vector (  ) with right hand. Direction of screw is direction of third vector (v). Turn to second vector (r). Direction of vectors

So C = (iA x + jA y ) x (iB x + jB y ) = iA x x (iB x + jB y ) + jA y x (iB x + jB y ) = i x i A x B x + i x j A x B y + j x i A y B x + j x j A y B y  A y = Asin  A x = Acos  A B C = A x B Vector Product A = iA x + jA y B = iB x + jB y C= ABsin  So C = 0 + k A x B y - kA y B x + 0 = 0 - k ABsin  now i x i = 0 j x j = 0 i x j = k j x i = -k

Is  a vector? However  is a vector! Rule for adding vectors: The sum of the vectors must not depend on the order in which they were added. Rule for adding vectors: The sum of the vectors must not depend on the order in which they were added.

This term is the tangential accel a tan. (or the rate of increase of v) The centripetal acceleration of circular motion. Direction to centre a and  r  Since  = v/r this term = v 2 /r (or  2 r)  v v Relating Linear and Angular variables

Total linear acceleration a The acceleration “a” of a point distance “r” from axis consists of 2 terms: Tangential acceleration (how fast v is changing)  a and  a =  r & v 2 /r Central acceleration Present even when  is zero! r Relating Linear and Angular variables a a

CM g  The whole rigid body has an angular acceleration  The tangential acceleration a tan distance r from the base is a tan  r At the CM: a tan  L/2, But at the CM, a tan = g cos   (determined by gravity) The tangential acceleration at the end is twice this, but the acceleration due to gravity of any mass point is only g cos   The rod only falls as a body because it is rigid The Falling Chimney L and at the end: a tan =  L gcos  ………..the chimney is NOT. ………..the chimney is NOT.

Kinetic Energy of a rotating body

It is clearly NOT ½ MV 2 cm since V cm = 0 What is the KE of the Rotating body? 1/2 MV cm 2 ??  cm

 m1m1m1m1 K rot = ½m 1 v ½m 2 v 2 2 +½m 3 v But all these values of v are different, since the masses are at different distances from the axis. However  (angular vel.) is the same for all. We know that v =  r. m2m2m2m2 V1V1V1V1 m3m3m3m3 K rot = ½  2  m i r i 2 Kinetic Energy of Rotation So that K rot =½m 1 (  r 1 ) 2 +½m 2 (  r 2 ) 2 +½m 3 (  r 3 ) 2 + =  ½m i r i 2  2 V3V3V3V3 V2V2V2V2

Where I is Rotational Inertia or Moment of Inertia of the rotating body K rot = ½ I  2 So K rot = ½ I  2 K rot =½ I  2 I=  m i r i 2 (compare K trans = ½ m v 2 ) K rot =½m(  r 1 ) 2 +½m(  r 2 ) 2 +½m(  r 3 ) =  ½m i r i 2  2 = ½  2  m i r i 2

I=  m i r i 2 Rotational Inertia “I” is the rotational analogue of inertial mass “m” For rotational motion it is not just the value of “m”, but how far it is from the axis of rotation. The effect of each mass element is weighted by the square of its distance from the axis The further from the axis, the greater is its effect.

K rot = ½ I  2 The bigger I, the more KE is stored in the rotating object for a given angular velocity A flywheel has (essentially) all its mass at the largest distance from the axis.

Some values of rotational inertia for mass M M R I=  m i r i 2 = MR 2 Mass M on end of (weightless) rod of length R

R I=  m i r i 2 =1/2 MR 2 + 1/2 MR 2 = MR 2 2 Masses M/2 on ends of (weightless) rod of length 2R (dumbell of mass M) M/2 Same as mass M on end of rod of length R... MR 2 Some values of rotational inertia for mass M

I=  m i r i 2 =  m i R 2 = MR 2 Mass M in a ring of radius R Same as mass M on end of rod, Same as dumbell... MR 2 R Some values of rotational inertia for mass M

mass of the rod M =  L L I =  m i r i 2 mass M Rotation axis x Rotational Inertia of a thin rod about its centre Linear density (kg/m) For finite bodies thickness dx M

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