1. The circumference of a circle equals 2 πR.  In Fig. 2 below,
   s equals one-quarter of a circle, or s = 2 πR/4 = πR/2.
   In radians Θ= s/R = ( πR/2)/R = π/2.  In Fig. 2,  Θ= 90^o.
   Thus π/2 radians = 90^o, or π radians = 180^o.
   
   
   
   
2. In general, (angle in radians)/(angle in degrees) = π/180^o.
   
   

1. Speed v at any point = the magnitude of the velocity tangent to the path at that point.
   
   
2. For circular motion Period T = time for complete rotation = 2 πr/v.
   
   
3. The frequency f = reciprocal of period = 1/T = v/2 πr.
   
   
4. The angular velocity ω (omega not w) = the angle turned through per second or for a varying angular velocity = dΘ/dt.
   
   In Fig. 1, Θ= (1/R)(s)            (Equation 1)
   
   Taking the derivative with respect to t on both sides of Eq. 1,
   
   dΘ/dt= (1/R)(ds/dt)      (Equation 2)
   
   Recognizing dΘ/dt as the angular velocity ω and ds/dt as the linear velocity v gives us the relation between ω and v.  ω = v/R.  Since the frequency f = v/2 πR or v = 2 πRf,  ω = (2 πRf)/R = 2 πf= 2 π/T.
   Unit of ω is s^-1.
   
   
5. Angular acceleration α is the rate of change of angular velocity.
   α= dω/dt.  Its unit is s^-2.
   
   For constant angular acceleration,
   
   
   1. 
      
      
   2. 
      
      
   3. ω² = ω_o² + 2α(Θ- Θ_o)
      Linear analogy, v² = v_o² + 2a(x – x_o)
      
      
6. Sample problems in 107 Problem Set for Angular Motion: 3-5.
   
   

1. All points on the object rotate with the same angular velocity ω.
   
   
2. For every mass m₁, m₂ m₃, etc. that makes up the body,
   v₁ = ωr₁,  v₂ = ωr₂,  v₃ = ωr₃, etc.  (See Fig. 3 below)
   
   
   
   
3. K = 1/2 m₁v₁² + 1/2 m₂v₂² + 1/2 m₃v₃² + + + +
      = ₁^NΣ(1/2) m_iv_i²
   K = 1/2 m₁r₁²ω² + 1/2 m₂r₂²ω² + 1/2 m₃r₃²ω² + + +
      = ₁^NΣ(1/2) m_ir_i²ω²
   Define Moment of inertia I = ₁^NΣ m_ir_i^2.  Then K = (1/2)Iω² .
   Compare with K = 1/2 mv² for translation.
   
   
4. If distribution of mass is continuous then integrate to find moment of inertia
   
      
   
   
   

1. The mass per unit area of the disk σ = M/ π R². The area of a thin ring of radius r, thickness dr, dA = 2πr dr. The mass of a thin ring of radius r, thickness dr = dm = (mass per unit area)dA = (M/ π R²)(2πr dr).
   
   
2. The moment of inertia of a disk about its center of mass is:
   
   
   
   
   

1. Mass/Length = M/L
   
   
2. Mass of length dx = dm = (Mass/Length)dx = (M/L)dx
   
   
3. dI = dm x² = (M/L) x² dx
   
   
   
   
   
4. About center of mass (Fig. 6a below),
   
   
   
   
5. About one end (Fig. 6b above),
   
   
6. Parallel Axis theorem: I(parallel) = I(CM) + Md², where d is the distance between the axes
   
   For rod about one end I = ML²/12 + M(L²/4) = ML²/3
   
   

1. At the top,

   E_top = U + K = Mgh + 0

2. At the bottom since the sphere has both translational and rotational kinetic energy,

   E_bottom = U + K = 0 + (1/2 Mv² + 1/2 I_CMω²)

3. Since ω = v/R,

   E_bottom = 1/2 Mv² + 1/2 I_CM(v²/R²)

4. From conservation of energy,

   Mgh = 1/2 Mv² + 1/2 (I_CM/R²)v² or v² = 2gh/(1 + I_CM/MR²)

5. Special Cases
   
   
   1. For hoop, I_CM = MR², v = (gh)^1/2
   2. For disk or cylinder, I_CM = MR²/2, v = (4gh/3)^1/2
   3. For sphere, I_CM = 2MR²/5, v = (10gh/7)^1/2
      
      

1. τ = dL/dt = d(Iω)/dt = Iα
2. Work = W = (Iα)Θ = τΘ
   
   

1. If no net external torque acts on a system, its total angular momentum L remains constant.  τ = dL/dt.  If τ = 0, dL/dt = 0,
   and L remains constant.
   
   
2. Examples
   
   
   1. Person on rotating table
      
      
   2. Planetary motion
      
      
   3. Gyroscope
      
      
      
      
      1. Wheel spins clockwise as viewed from support
         L is to the right.  (Fig. 10a above)
         
         
      2. The torque due to the weight mg is τ = r x mg points into the paper. (Fig. 10b above)
         
         
      3. As seen from above, L precesses slowly counterclockwise. (Fig. 10 c and d above)
         
         L + ΔL = L + τ Δt,  where τ = dL/dt.
         d(L²)/dt = d(L ^. L)/dt = L ^. dL/dt + L ^. dL/dt = 2L ^. dL/dt =
         2L ^. τ = 2Lτcos L, τ
                  =2Lτcos 90^o = 0
         
         Since τ is perpendicular to L, there is no change in the magnitude of L, only change is in its direction.
         
         
      4. dΦ = τdt/L  or  Ω = dΦ/dt = τ/L = mgr/Iω.
         Wheel spins at rate ω and precesses at rate  Ω.
         
         

1. the moment of inertia of the apparatus,
2. The angular acceleration produced by the child,
3. how long it will take for the apparatus to make one complete revolution,
4. at the time for (3)
   1. the angular velocity, and
   2. the kinetic energy of the apparatus, and
5. the work done by the child.
6. Find the direction and magnitude of ΔL/ Δt and the torque.
   
   

1. Treating the two carts as point particles,
   I = 50 kg(2 m)² + (50 kg)(4 m)² = 1000 kg-m².
   
   
2. τ = r x F
   τ= (4 m)(10 πN) sin 90^o = 40 πN-m = Iα = 1000 kg-m²α
   α= 0.04 πs^-1
   
   
3. Angle turned through in one complete revolution = 2 π = 1/2 αt².
   t = [(4 π)/α]^1/2 = [(4 π)/4 πx 10^-2 s^-2]^1/2 = 10 s
   
   
4. 1. (a) ω = ω_o + at = 0 + (4 πx 10^-2 s^-2)(10 s) = 0.4 πs^-1
   2. (b) K = 1/2 Iω² = 1/2 (1000 kg-m²)(0.4 πs^-1)² = 80 π² J
      
      
5.  W = τΘ= (40 πN-m)( 2 π) = 80 π² J = K_f – K_I = 80 π² J – 0
   
   
6. Since ω_o = 0,  L_o = 0.  Final L = Iω.  Since the apparatus rotates counterclockwise, ω is out of the page.  After one revolution,
   L = (1000 kg-m²)(0.4 πs^-1) = 4 πx10² m-kg/s.
   ΔL/ Δt = (4 πx10² m-kg/s)/(10 s) = 40 π m-kg/s² = τ out of page.
   
   
   

1. For translation,

   F_net = ma
   mg sin Θ - f = ma          (Equation 1)

   
   
2. For rotation about the center of mass, neither N nor mg produces a torque about the center of mass because their line of actions pass through the center of mass.

   τ_CM = I_CMα
   fR sin 90⁰ = (2/5 mR²)α   or
   since  α = a/R,
   f = 2/5 ma                   (Equation 2)

   Substituting (2) into (1):
   
   mg sin Θ - 2/5 ma = ma   or
   a = 5g sinΘ/7 v² = v₀² + 2as = 0 + (10/7)g(s sinΘ) = (10/7)gh,
   v = (10gh/h)^1/2, as we found before using conservation of energy.
   
   
   

Σ F = 0

Στ = 0

1. The forces acting on the beam are:
   
   
   1. The gravitational force of the earth on the beam,
      its weight = 200 N acts at the center of the uniform beam.
      
      
   2. The force of the man on the beam = 600 N
      
      
   3. The tension T in the rope
      
      
   4. The force R exerted by the wall at the pivot
      
      
2. The torques acting on the beam about the axis at the pivot point are:
   
   
   1. Torque due to R is zero because it acts at the pivot point. Any force that passes through the axis produces no torque (r = 0).
      
      Note: I chose the axis at the pivot point to eliminate one of the unknowns in the torque equation, leaving only T as an unknown.
      
      
   2. The torque due to the tension in the rope = 8.0 m(T) sin 53^o
      out of the page.
      
      
   3. The torque due to the force of the man = 2.0 m(600 N)
      into the page
      
      
   4. Since it is a uniform beam, the weight of the beam acts at the center.
      
      r = 4 m for the beam weight.
      Torque = 4.0 m(200 N) into the page.
      
      
   5. Στ = 0
      
      Taking out of the page as positive:
      

      +8.0 m(T) sin 53^o - 2.0 m(600 N) - 4.0 m(200 N) = 0
        8.0 m(T)(0.80) = 2000 N-m.    T = 313 N

      
      
3. The vector sum of the forces must = 0.
   
   Change this two-dimensional problem into two one-dimensional problems, by taking the components of the tension T and the force of the wall at the pivot R into X and Y components.
   
   As shown in Fig. 12b,

   T_x = T cos 53^o
   T_y = T sin 53^o,
   R_x = R cos Θ,   and
   R_y = R sin Θ

   ΣFx = 0	ΣFy = 0
   R cos Θ - T cos 53o = 0	R sin Θ + T sin 53o – 600N – 200N = 0
   R cos Θ = T cos 53o	R sin Θ = - T sin 53o + 600 N + 200 N
   R cos Θ = 313(0.6)	R sin Θ= -313 N(0.8) + 800 N
   R cos Θ = 188 N  (Equation 1)	R sin Θ= 550 N     (Equation 2)

   
   
   Dividing Eq.2 by Eq. 1:

   tan Θ= 550 N/188 N = 2.92.    Θ = 71.1⁰ R.
   R sin Θ= 550 N.
   R = 550 N/sin Θ.
   R = 550 N/sin 71.1^o = 581 N.

   
   

1. Magnitude of C = A x B is C = |A||B|cos A, B.
   
   
2. Point the fingers of your right hand in the direction of A with your palm toward B.  Rotate your palm into B.  Your thumb points in the direction of C.  In Fig. 14 below, C is out of the page.
   
   
   
   

1. i x i = |i||i|sin i, i = |1||1|sin 0^o = 0
   j x j = |j||j|sin j, j = |1||1|sin 0^o = 0
   k x k = |k||k|sin k, k = |1|1|sin 0^o = 0
   
   
2. Magnitudes and Directions
   
   
   1. Magnitude: |i x j| = |i||j|sin i, j = |1||1|sin 90^o = 1 = |j x i|
      Direction: i x j = k     j x i = - k
      
      
   2. Magnitude: |i x k| = |i||k|sin i, k = |1||1|sin 90^o = 1 = |k x i|
      Direction: i x k =- j      k x i =j
      
      
   3. Magnitude: |j x k| = |j||k|sin j, k = |1||1|sin 90^o = 1 = |k • j| Direction: j x k = i     k x j = - I