It’s just a new notation. ROTATION “X” is q, “v” - omega It’s just a new notation. The change, you see, Is due to the Radius of rotation

ROTATION Try to remember these when working with rotation: Always use radian measure One radian (rad) = 57.30 1 revolution (rev) = 360o = 2p rad Angular displacement is not zero after one rotation The angular velocity vector is perpendicular to the direction of motion and along the axis of rotation (right hand rule) Counterclockwise rotation is positive (“clocks are negative”)

RIGHT HAND RULE FOR ROTATING OBJECTS Curl your right hand about the object with your fingers pointing in the direction of motion. Your extended thumb will point in the direction of the angular velocity vector

TRANSLATING ROTATION USING TRANSLATION Only apply to constant acceleration

EXAMPLE ONE (a) w(2) = 4 rad/sec (b) aave = 12 rad/s2 The angular position of a point on the rim of a rotating wheel is given by: q = 4.0t – 3.0t2 + t3, where q is in radians and t is in seconds. (a) What is the angular velocity at t=2s? (b)What is the average angular acceleration for the time interval t = 2 to t=4 seconds? (c) What is the instantaneous angular acceleration at 4 seconds? (d) What is the angular displacement of the point for the time interval t=2 to t=4 seconds? (e) How many rotations will the wheel have gone through? (a) w(2) = 4 rad/sec (b) aave = 12 rad/s2 (c) a(4) = 18 rad/s2 (d) Dq = 28 radians (e) 4.46

EXAMPLE TWO Calculate the total rotational kinetic energy of three small spheres that revolve around a vertical axis at an angular velocity of 6 rad/s and the following masses and radii: m1 = 0.5kg r1 = 0.25m m2 = 0.4kg r2 = 0.3m m3 = 0.6kg r3 = 0.4m m2 m1 r1 r2 r3 m3

ROTATIONAL KINETIC ENERGY The combined kinetic energies of all of the pieces, makes up the total rotational kinetic energy. Recall v = wr and w is the same for each piece (m) and Krot =1/2m(w2r2) The term mr2 for each piece is called the rotational inertia or moment of inertia symbolized by ‘I’ with units of kgm2 Finally, Krot = ½ Iw2 For a large rigid object – imagine it is made of many small masses.

EXAMPLE TWO - SOLUTION Ktotal =K1 + K2 + K3 m1 r1 r2 r3 m3 Ktotal =K1 + K2 + K3 K = ½ w(m1r12+m2r22 +m3r32) K = 0.49 J

LIVE FOR THE MOMENT OF INERTIA

MOMENT OF INERTIA Standard form for moment of inertia: For objects rotating about A point other than the c.o.m. Use the parallel axis theorem Calculate the rotational inertia for a thin rod (like a meter stick) being swung by one end To calculate the rotational inertia for a thin hoop about a central axis – break the hoop into small masses (dm) and sum them I = r2M where M = total mass

FORCE AND ROTATION Instead of an external force causing linear motion, a force exerted at some distance from an axis of rotation causes a TORQUE (picture opening a door) Torque is the cross product of force and radius (t =r x F, or t = (r)(Fsinq)) Only the component of the force perpendicular to the radius vector causes a torque TORQUE IS NOT THE SAME AS WORK!!!

NEWTON’S SECOND LAW AND ROTATION For straight line motion: Fnet = m a Since: t = F r, a = a r, and I = m r2 We get: t = Fr = mar = m (ar)r = mr2a That results in: tnet = I a NOTE: Torques causing clockwise rotation are negative

EXAMPLE THREE tnet = t1 – t2 t1=F1r1sinq1 t1=F2r2sinq2 Calculate the net torque about point O for the object below if F1=4.2N, F2=4.9N, q1=750, q2=60o, r1=1.3m, and r2=2.15m tnet = t1 – t2 t1=F1r1sinq1 t1=F2r2sinq2 tnet = -3.85 N m

EXAMPLE FOUR (a) a = .06m/s2 (b) TM=4.87N,Tm=4.54N (c) a = 1.2rad/s2 For the system shown: M=.5kg, m=.46kg, rpulley=0.05m and when released from rest, M falls .75m in 5 seconds Calculate the following: (a) acceleration of the blocks (b) tension in the cords (c) angular acceleration of the pulley (d) rotational inertia of the pulley (a) a = .06m/s2 (b) TM=4.87N,Tm=4.54N (c) a = 1.2rad/s2 (d) I = 1.38x10-2 kg/m2

And what do I want in return? Just your best effort every day… Now it’s your ‘turn’ And what do I want in return? Just your best effort every day… That’s all!