Cheatography

# MidTerm 3 by brandenz1229

### Chapter 9

 Angular Velocity and Accele­ration Θ = angle (radians) s = length r = radius 90o = π/2 rad Θ = (s/r) s = r ⋅ Θ 1 rad = (360o / 2π) = 57.3o 180o = π rad Angular Velocity (1st Deriva­tive) ω = (Θ`f`- Θ`i`) / (t`f`-t`i`) ω = "­vel­oci­ty" 1 rev/s = 2π rad/s 1 rev/min = 1 rpm = 2π/60 rad/s Angular Accele­ration (2nd Deriva­tive) α = (ω`f`- ω`i`) / (t`f`-t`i`) α = "­acc­ele­rat­ion­" Rotation w/ Constant Angular Accele­ration α`f` = (ω`f`- ω`i`) / (t - 0) α`f` = constant ω`f` = ω`i` + α`f` ⋅ t Θ`f`- Θ`i` = 1/2(ω`i`+ω`f`) ⋅ t Θ`f` = Θ`i`+ (ω`i`⋅ t ) + 1/2 (α`f`⋅ t2) ω`f`2 = ω`i`2 + 2⋅α`f`(Θ`f`- Θ`i`) Relating Linear and Angular Kinematics K = 1/2(m ⋅ v2) Linear Speed in Rigid-Body Rotation s = r ⋅ Θ Linear Speed v = r ⋅ ω Linear Accele­ration in Rigid-Body Rotation a`tan` = r ⋅ α Centri­petal Component of Accele­ration a`rad` = (v2/r) = ω2⋅r Energy in Rotational Motion K`E`: 1/2⋅m⋅v2 = 1/2⋅m⋅r2⋅ω2 K = 1/2⋅m⋅r2⋅ω2 I = m⋅r2 Gravit­ational Potential Energy for an Extended Body U = M⋅g⋅y`cm` Moment of Inertia I`p`=I`cm`+Md2

### Chapter 9 Cont:

 Rotational Kinetic Energy K = Joules K = 1/2⋅I⋅ω2 R = Radius M = mass pivoted about an axis Perpen­dicular to the Rod I = (M⋅L2) / 3 Slender Rod (Axis Center) I = 1/12M⋅L2 Slender Rod (Axis End) I = 1/3M⋅L2 Rectan­gular Plate (Axis Center) I = 1/12M⋅(a2+b2) Rectan­gular Plate (Axis End) I = 1/3M⋅(a2) Hallow Cylinder I = 1/2M(R`i`2+R`f`2) Solid Cylinder I = 1/2MR2 Hollow Cylinder (Thin) I = MR2 Solid Sphere I = 2/5MR2 Hollow Sphere (Thin) I = 2/3MR2

### Chapter 11: Equili­brium and Elasticity

 1st Condition of Equili­brium (at rest) ΣF = 0 2nd Condition of Equili­brium (nonro­tating) Στ = 0 Center of Gravity r`cm` = (m`1`⋅ r`1`) / m`1` Solving Rigid-Body Equili­brium Problems ΣF`x` = 0 1st Condition ΣF`x` = 0 ΣF`y` = 0 2nd Condition (Forces xy-plane) Στ`z` = 0 Stress, Strain, and Elastic Moduli Stress = Force Applied to deform a body Strain = how much deform­ation Hooke's Law (Stress / Strain) = Elastic Modulus A = Area F = Magnitude of Force Tensile Stress F / A 1 Pascal = Pa = 1 N/m2 1 psi = 6895 Pa l = length 1 Pa = 1.450 ⋅ 10-4 Tensile Strain (l`f`- l`i`) / (l`i`) Young Modulus (Tensile Stress) / (Tensile Strain) Pressure p = F (Force Fluid is Applied) / A (Area which force is exerted) Bulk Stress (p`f` - p`i`) Bulk Strain (V`f`- V`i`) / (V`i`) Bulk Modulus Bulk Stress / Bulk Strain

### Chapter 10: Dynamics of Rotational Motion

 Torque F = Magnitude of F || || = Magnitude Symbol τ = F⋅l = r⋅F⋅sinΘ = F`tan`r L = lever arm of F τ = ||r|| x ||F|| Torque and Angular Accele­ration for a Rigid Body Newtons 2nd Law of Tangential Component F`tan` = m`1`⋅a`1` Rotational analog of Newton's second law for a rigid body Στ`z` = l⋅α`z` `z` = rigid body about z-axis Combined Transl­ation and Rotation: Energy Relati­onships K = 1/2M⋅v2 + 1/2⋅I⋅ω2 Rolling without Slipping v = R⋅ω Combined Transl­ation and Rotation: Dynamics Rotational Motion about the center of mass Στ`z` = l⋅α`z` Work and Power in Rotational Motion F = M ⋅ a When it rotates from Θ`i` to Θ`f` W = ∫ (Θ`f` to Θ`i`) τ`f` dΘ When the torque remains constant while angle changes W = τ`f`(Θ`f` to Θ`i`) Total WorkDone on rotating rigid body W = 1/2(ω`f`2) - 1/2(ω`i`2) Power due to torque on rigid body P = τ`z`⋅ ω`z` Angular Momentum L = r x p (r x m ⋅ v) Angular Momentum of a Rigid Body L = m`i`⋅r`i`2⋅ω

### Chapter 11: Equili­brium and Elasticity (cont.)

 F = Force acting tangent to the surface divided by the Area Shear Stress F / A h = transverse dimension [bigger] x = relative displa­cement (empty) [smaller] Shear Strain x / h