First, we derive the equations governing rotational motion from the definitions of basic quantities of rotational motion in 8.1-8.4. Note the parallelism between rotational motions and translational motions.

Next, we solve some problems as examples. Always start solving a problem using symbols to set up the equations you need to use and plug in numbers at the end.

8.1 Angular Displacement

We will consider circular motion only in this chapter to minimize mathematical complications.

A particle moves in a circular orbit. Two ways to describe such a motion:

• We can describe its displacement by the arc-length l along the circumference, and the velocity V tangent to the circle at the location of the particle.
  
  
• We can also describe its motion by the angular displacement q, and the angular velocity w as specified in Figure 8.2(a), where r is the radius of the circle
  

Figure 8.2 shows the geometrical relationships among arc-length l (in meters), the angular displacement q, (in radians) and the radius r (in meters) of a circle. It is self evident that l is proportional to the angular displacement q. This leads to the equation given below:

Angular displacement q is related to linear displacement l by equation (1).

Conversion between radians to degrees is given by 2p (radians) = 360 (degrees)

Radian is a dimensionless unit, i.e., a pure number. This is seen from C = 2pR where C is in meters and R is also in meters so that p in rad must be dimensionless.

8.2 Angular Speed

Average angular speed is defined by:  wav = Dq/Dt.

From (1) rDq = Dl since r is constant on a circle. Dividing both sides by Dt, we show or derive

Angular speed wav is related by linear or tengential speed Vav in equation (2)

EX: P33 and Figure P33 on page 279 The pulley on the left of Fig. P33 is 0.6 in diameter and rotates at 1.0 rpm. It is attached by a twisted belt to the 0.2-m-diameter hub of a compound pulley whose outer diameter is 0.8 m. If the driver turns clockwise, find the velocity of the suspended body.

Solution: fa = 1.0 rpm = 1/60 rps    and    Va = (2pRa) fa -------- (1) Va = Vc = (2pRc) fc ----------- (2) (1) and (2) leads to Rafa = Rcfc     or    fc = fa (Ra/Rc) fc = fd     and    Vd = (2pRd) fd = (2pRd) fc = (2pRd) fa (Ra/Rc) Vd = (2pRd) (Ra/Rc) fa = (6.28x0.4)x(0.3/0.1)x(1/60) = 0.125 m/s The direction of Vd is upward because pulley A rotates clockwise causing the compound pulley C and D to rotate counter-clockwise.

8.3 Angular Acceleration

Average angular acceleration is defined by: aav = Dw/Dt.

From this definition we show aav = Dwav/Dt = r Dwav / rDt =  DVav / rDt = aav / r

Angular acceleration aav is related to linear or tangential acceleration aav in equation (3)

8.4 Equations of Constant Angular Acceleration

From the definitions of angular acceleration and angular speed, derive the following equations (do them on your own to practice derivation of equations):

HW Chap 8 (Part-I): P5, P19, P27, P41, P51

8.5 Torque: defined by t = r.F.sinq = r.F^ = r^ .F

Torgue is responsible for rotational motion just as force is responsible for translational motion.

EX: Study example 8.9 on page 252 on your own. Ask questions if any.

8.6 Equilibium Conditions: (1) SF = 0 for Translational Equilibrium
                          (2) St = 0 for Rotational Equilibrium

Ex: A 8000-N car is stalled one-quarter of the way across a bridge (see Figure P65). Compute the additional reaction forces, Fa and Fb, at supports 'a' and 'b' due to the presence of the car at 'c'. Take the length of the bridge is L. From (1)        Fa + Fb = 8000 N From (2)        8000 Lca - Fb L = 0,                8000 Lcb - Fa L = 0,

If Lcb = L/4, solve these equations to show that Fa = 2000 N and Fb = 6000 N.

8.8 Torque and Rotational Inertia:

If a rigid body consists of several discrete mass elements, the net torque on the body is the sum of torques on all mass elements:

t = Sti = Sr.F.sinq = Sr.F^ = Sr^ .F = Sr^ .ma = Sr^ m.(r^ a) = (Sm.r^ **2) a = I. a

If the mass elements in a body are not discrete, but form a continum, then the summation sign turns into integration sign in calculus. Table 8.3 show the results of rotational inertia for various body geometries by means of integration. We will use these expressions without deriving them.

Ex: 8.14

From Figure 8.27 one can set up the solution to the problem without using any numerical values. When both rotational and translational motions are involved, we must use both torque equation and force equation, i.e., Let's denote M = mass and R = radius of the cylinder and FT = tension in the rope, we have

     t = FT.R = I.a = [(1/2).M.R**2].a         

     F = m.a = m.R.a = mg - FT   

The unknowns are a and FT. Solve these two unknowns from the above two equations. Show that

         a = mg/[m.R + (1/2)MR]

           FT = mg - mg/[1+ (1/2)M/m] = mg[1-2m/M]

If you need the linear acceleration, you can get it from a = Ra

EX: A hoop of mass m and radius R rolls without slipping down an incline at an angle q.  Write an expression for the acceleration of its center of mass.

8.9 Rotational Kinetic Energy:

Derive the above result:

A rigid body rotates about an axis. Each mass element m in the body moves with the velocity V so that the kinetic enrgy of the rotating body is given by

       KER = S(1/2) mV **2 = (1/2) S m (R w)**2 = (1/2) S m R**2.w**2 = (1/2)I.w**2

A rigid body in both translational and rotational motions, its kinetic energy is given by

      KE = KET + KER = (1/2)mV**2 + (1/2)Iw**2

EX:

A hoop of mass m rolls down an inclined plane from a height h. If it begins at rest,show that its final speed is given by V**2 = gh. (P125 on page 287)

                     KER + KET = PE    or    (1/2)Iw**2 + (1/2)mV**2 = mgh --------- (1)

                     (1/2)Iw**2 = (1/2)mR**2(V/R)**2 = (1/2)mV**2 ----------------- (2)

(2) into (1):     (1/2)mV**2 + (1/2)mV**2 = mgh

                     mV**2 = mgh     or    V**2 = gh

8.10 Angular Momentum

Definition of Angular Momentum

A particle of mass 'm' moving at a speed 'V'. Its angular momentum 'Lo' with respect to a point 'O' is defined by    Lo = r^  P = r^ m V = r^ m (r^ w)            = (mr^ **2)w   where r^ = r sinq

For an extended solid body, made up of many particles, rotating about an axis through point 'O', the angular momentum of the body is the summation of angular momentum of individual particles.

         Lo = S r^  P = S r^ mV = S r^ m(r^w) = (S mr^ **2)w = Iw

EX: A uniform disk of mass 800 kg and radius 0.5 m is rotating around its central symmetry axis at a rate of 60 rpm. Determine its angular momentum.

     I = (1/2)mR**2 = 400 x 0.25 = 100 kg.m.m     

     w = 2p x 60 radian/m = 2p rad/s

   L = Iw = 100 x 2p kg.m**2/s

8.11 Conservation of Angular Momentum

If the net torque acting on the system is zero, i.e., t = 0, then 

EX:  When the nuclear fuel runs out in a large star, it collapses into a tiny neutron star. Our Sun spins at a rate of about once every 27 days. Its radius is 6.9 x 10**5 km. What would be the spin rate if our Sun collapses into a neutron star of 20 km radius?

Conservation of angular momentum:  Ii wi = If wf ---------------------------------- (1)

     w = 2pf and I = (2/5) m R**2 for a solid sphere (given in Table 8.3)

Equation (1) reduces to       fi Ri**2 = ff Rf**2

     fi = 1/(27x24x60x60) rev/s = 4.3 x 10**-7 rev/s

     ff = fi (Ri /Rf)**2 = 4.3 x 10**-7 x (6.9 x 10**5 / 20)**2 = 526 rev/s

HW Chap 8 (Part-II): P71, P77, P107, P135