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Created by Hayden Tornabene
almost 9 years ago
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Question | Answer |
Blocks on Ramps: Force Parallel and Force Perpendicular |
Fg,par=mgsinθ
Fg,perp=mgcosθ |
Force of Friction (Normal Force) |
Ff=μmgcosθ
N=mgcosθ |
Kinematic Equations |
x(t)=v0xt+x0
y(t)=−12gt2+v0yt+y0
Only use kinematics if you need to know the explicit time dependence of a system. Energy considerations are faster otherwise.
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Initial and Final Velocity Equation |
v2f−v2i=2aΔx |
Radial Acceleration for Uniform Circular Motion |
a=v2r |
Centripetal Force for Uniform Circular Motion |
F=mv2r
Does not tell you what kind of force acting on body, just that a body moving in uniform circular motion has this force.
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Types of Energy |
Translational Kinetic: 12mv2
Rotational Kinetic: 12Iω2
Gravitational Potential: mgh
Spring Potential: 12kx2 |
Potential Energy Line Integral |
ΔU=−∫baF⋅dl
For any conservative force - Integral is indirection of the force vector.
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Gravitational Force and Gravitational Potential |
Fgrav=Gm1m2r2ˆr
U(r)=GmMr |
Potential Energy and Force Relationship |
F=−∇U |
Linear Velocity of Rolling without Slipping |
v=Rω
Questions regarding which arrives with slower linear velocity are NOT questions of which arrive first. Need kinematics.
Ex. energy expression for a sphere with mass m and radius r.
mgh=12mv2+12(25mr2)ω2
REMEMBER: ω=vr |
Work Energy Theorem - In terms of Kinetic Energy - In terms of Force |
Einitial+Wother=Efinal
W=ΔKE
W=∫F⋅dl |
Linear Collisions Conservation of Momentum Two balls collide, M on m with M scattering at θ with initial velocity V with the final velocity equal for both at v. What is the scattering angle of ϕ of m? |
MV=Mvcosθ+mvcosϕ (parallel to initial)
0=Mvsinθ+mvsinϕ
(perpendicular to initial)
ϕ=sin−1(−Mmsinθ)
Using limiting cases, triaging is essential. Finding phi analytically takes time.
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Angular momentum of a point particle and for an extended body. |
→L=→rx→p
→L=I→ω |
Torque (Angular Force) |
→τ=→rx→F |
Scalar Analogues of →p=m→v and →F=d→pdt | Angular Momentum: L=Iω Torque: τ=dLdt The vector angular momentum →L is generally parallel to →ω with the angular momentum determined by the right hand rule. |
Moment of Inertia (General and Extended Objects) |
I=mr2
I=∫r2dm for dm = ρdV and
ρ=Ar3
Compute the integral to find the constant A and then insert into the extended objects integral.
Ex. ρ(x)=Ax2
M=∫L0ρ(x)dx=13AL3→A=3ML3 |
Parallel Axis Theorem |
I=ICM+Mr2
For penny through center, I=12MR2
For penny axis through edge, I=12MR2+MR2
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Center of Mass Displacement from Origin |
→rCM=∫→rdmM
Must solve three integrals, one for each dimension. Remember to select a volume integral and corresponding coordinate system that capitalizes on symmetry.
For a system of point masses:
→rCM=Σi→rimiM |
Lagrangian Function |
Scalar Function
L(q,˙q,t)=T−U
for T kinetic energy, U potential energy, and q the generalized coordinate describing degrees of freedom.
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How do you compute the correct expression for T in the Lagrangian? | 1. Write down expressions for Cartesian coordinates in terms of chosen coordinates q. 2. Differentiate x,y,z to get ˙x,˙y,˙z 3. Form kinetic cartesian energy, T=12m(˙x2+˙y2+˙z2) for point particle, T=12Iω2 for an extended object. For extended object, must express ω in terms of ˙q, but this is easy as the two are generally equal. See page 20 in notebook for excellent example. |
Euler-Lagrange Equations |
ddtδLδ˙q=δLδq
One equation for each degree of freedom, q.
Remember that ddt is a total time derivative, not a partial derivative. I.e. ddtm˙x=m¨x
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Momentum Conjugate to q (Lagrangian) |
δLδ˙q
Iff the Lagrangian is independent of a coordinate q, the corresponding conjugate momentum quantity, δLδ˙q, is conserved.
Quantities whose time derivatives are zero are conserved quantities.
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Hamiltonian |
H(p,q)=Σipi˙qi−L
for pi=δLδ˙qi
If the potential energy is not explicitly dependent on velocities of time,
H = T + U
Iff the Hamiltonian is independent of a coordinate 1, the corresponding conjugate momentum p is conserved.
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Hamiltonian for Particle Moving in One Dimension |
H=T+U=p2x2m+U(x) |
Hamiltonian for Particle Moving in 2D Polar |
H=p2r2m+p2θ2mr2 |
Hamilton's Equations |
Scalar function encoding equations of motion:
˙p=−δHδq
˙q=δHδp |
Lagrangian for a Simple Pendulum |
Take corresponding derivatives - e.g. ˙x=lcosθ˙θcosϕ−lsinθsinϕ˙ϕ
Plug the ˙x,˙y,˙z into the expression for T=12m(˙x2+˙y2+˙z2)
Simplify and put into L = T - U for U=mgz=−mglcosθ
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Conservation of Conjugate Momentum Example |
For the Lagrangian independent of ϕ, pϕ is conserved.
Therefore the following quantity is conserved:
pϕ=δLδ˙ϕ |
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