1-D Motion
2-D Motion
Forces
Work & Energy
Work/Power
Momentum/SOP
Center of Mass
&
Moment of Inertia
&
Moment of Inertia
Circular Motion
Torque
Angular Energy
&
Momentum
&
Momentum
Periodic Motion
Gravitation
100
This is the fourth kinematic equation not on the equation sheet and is also the beginning of the derivation that vf = 2 x vaverage.
Δx = ½ (v_o + v)Δt
100
This combination of 2-D kinematic equations is used to calculate the maximum range of a projectile.
Δx_max = [2v_0^2 (sintheta)(costheta)] / g
100
The force of gravity near the surface of the Earth is equal to the product of these quantities.
What is mass and the acceleration due to the force of gravity (g)?
F_g = W = mg
100
When calculating height in a situation when a pendulum is involved and the context is energy, use this equation that is always equal to the maximum height that the pendulum will rise above its equilibrium point.
h = L - Lcostheta
100
Potential Energy of conservative forces.
Derive the equation for Ugravity
Start with Fg
Fg = -(GMm) / (r2) & Fg = -dU/dx
Fg dx = -dU so -(GMm) / (r2) = -dU
∫ -(GMm) / (x2) dx = ∫ -dU
U = -GMm / x or U = -GMm / r
100
Equation for momentum.
p = mv
100
This is the equation for the center of mass for a rigid object.
r_(cm) = 1/(m_(Total))intr dm
100
These are the equivalencies for angular velocity.
omega = [d(theta)]/dt = v/r = 2pif = (2pi)/T
100
Torque is equivalent to these two equations.
Sigmatau = Ialpha
Sigmatau = (dL)/dt
100
These two equations are equivalent to angular momentum.
L = r x p = rp sintheta or rmv sintheta
L = Iomega
100
Angular Velocity of a Mass/Spring System.
Period of a Mass/Spring System.
Period of a Pendulum
ω= sqrt(k/m) = (Delta(theta))/(Delta t)=2T
T =2pi sqrt(m/k)
T=2pi sqrt(L/g)
100
Gravitational Force
F_g = (Gm_1m_2)/r^2
200
Kinematic Equation that contains acceleration, time and change in velocity.
v = v_o + aDeltat
200
When using the range equation, the change in this quantity must be equal to 0.
What is Δy?
200
These are the component vectors of weight if the object is not oriented aligned to the vertical y and horizontal x axis.
F_∥ = mgcostheta and F_⟂ = mgsintheta
200
Work can be expressed as the dot product of Force and displacement F · d. It can also be equal to this.
Positive work results in this effect to energy.
Negative work results in this effect to energy.
What is ΔE?
What is increase the energy?
What is decrease in energy?
200
Average Power is equal to these two equations.
Pav = F cos(theta) v
P_(avg) = W / (Deltat) = (Fd)/(Deltat) = (mgd)/(Deltat)
200
Impulse-Momentum Theorem can be written in both of these forms.
F_(external) = (dp)/dt = 0 = Sigmap_i = Sigmap_f
J = FDeltat = Deltap = p_f - p_i
200
This is the equation for Volumetric Mass Density.
ρ = m / ∀
200
Equations that translate:
1. translational velocity to angular velocity and
2. displacement to angular displacement
v = r𝛚
s = r𝜃
200
Parallel Axis Theorem
Use when asked for the moment of inertia around a new point when the moment of inertia for the object is already know.
200
Angular Momentum relationship to torque when the sum of the torques equals 0.
(dL) / dt = 0 = Sigma tau_(external)
200
Condition for Periodic Motion in Simple Harmonic Motion
(d^2x)/dt^2 = -omega^2x
200
Gravitational Potential Energy when height is very large.
U_g = -(Gm_1m_2)/r
300
Kinematic Equation that is used to find change in position.
Deltax = v_oDeltat + 1/2aDeltat^2
300
These are the two quantities that are constant in 2-D kinematics.
What is acceleration in the y-direction?
What is velocity in the x-direction?
300
Force of friction is equal to the product of these two quantities.
F_f = muF_N
300
Work is always equal to these two quantities.
What is Kinetic Energy and Force x Displacement?
300
Potential Energy of conservative forces.
Derive the equation for Potential Energy Us
Start with the Fs
Fs = -kx & Fs = -dU/dx
Fs dx = -dU so -kx dx = -dU
∫ -kx dx = ∫ -dU
U = 1/2 kx2
300
Momentum for a system of particles.
Sigmam_iv_i = p
300
This is the equation for Surface Mass Density.
σ = m / A
300
Relationship from angular acceleration to angular velocity and translational acceleration.
alpha = (domega)/dt
a_T = ralpha
300
Equation for moment of inertia.
I = 𝚺mr2
300
A rolling object of mass m starts from rest and rolls down an inclined plane of height h. Write the conservation equation for the object given the moment of inertia I, angular velocity 𝛚 and translational velocity v halfway down the inclined plane. Write in terms of the given variables and constants.
MEi = MEf
mghi = 1/2 mv2 + 1/2 I𝛚2 + mghf
300
Position of an object in Simple Harmonic Motion
y = Asin(omegat + phi)
300
Kepler's 3rd Law - The law of periods
T^2 = ((4pi^2)/(Gm_p))R^3
400
Kinematic Equation that contains the square of both velocities.
v^2 = v_o^2 + aDeltax
400
Acceleration in the x direction is always this.
What is 0 m/s2?
400
This is the equation used for Hooke's Law, which describes the force acting when a spring or rubber band is stretched or compressed.
F = -kx
400
W_(Friction) = DeltaME
only if this is also true.
No other forces are acting.
400
Conditions that work is done.
When an external force is applied in the same direction as object is moving.
Work = F*dcos𝜃
400
The following is a statement of this;
∑ p_i = ∑ p_f
What is conservation of momentum?
400
This is the equation for Linear Mass Density.
λ = m / L
400
Centripetal Acceleration
a_c = v^2/R
400
Moment of inertia of a particle in space.
I = mr2
400
A spinning disc of mass M and moment of inertia IA is rotating with angular velocity 𝛚. Another disc of mass m and moment of inertia IB is dropped onto the spinning disc. Write the conservation of momentum equation that represents this scenario in terms of M, m, I and 𝛚.
L = L'
IA𝛚M = IA𝛚M' + IB𝛚m' = (IA + IB)𝛚
400
Velocity and Acceleration of an object in Simple Harmonic Motion.
v = omega A cos(omegat + phi)
a = -omega^2Asin(omegat + phi)
400
Derive the escape velocity from a planet of mass M.
Hint: Use energy. Ei = Ef
Ei = Ef KEi + Ugi = KEf + Ugf
1/2 mrvri2 + (-GMEmr)/r = 1/2mrvrf2 + (-GMEmr)/r
1/2vri2 - GME / rE + GME / r = 1/2 vrf2
r goes to infinity and velocity final goes to 0
1/2 vri2 = GME / r
vescape = sqrt((2GME)/r
500
In order to use the kinematic equations, this quantity must be constant.
What is acceleration?
500
Maximum height equation.
Δy_max = [v_0^2 (sin^2theta)] / (2g)
500
These are alternative ways to think of
SigmaF = ma
F = (mdv)/dt
F = (dp)/dt
500
This equation is used when there is a conservative force, like a spring, and the potential energy associated with that force.
F_x = -(dU)/(dx)
500
Conditions that no work is done.
When a force is applied but the object doesn't move.
When a force is applied perpendicular to the direction of motion of the object.
Work = Fdcos𝜃
500
Derive the final v in a collision between a bullet m and a block M hanging from a chain.
Derive the equation for the maximum height hmax that the bullet/block mass reaches on the pendulum.
mv = (M + m)v'
v' = mv/(M+m)
1/2(M+m)(v')2 = (M+m)h
h = (v')2 / [2(M+m)]
500
Equation for parallel axis theorem.
Inewcm = Ioldcm + Mtotald2
500
Angular Kinematic Equations
omega^2 = omega_o^2 + alphaDeltatheta
Deltatheta = 1/2(omega_o + omega)Deltat
Deltatheta = omega_oDeltat + 1/2alphaDeltat^2
omega = omega_o + alphaDeltat
500
Maximum Velocity of the object in Simple Harmonic Motion
Maximum Acceleration of the object in Simple Harmonic Motion
v = omega A
a = -omega^2A
500
Time of an orbit of a satellite of mass m around the earth at a velocity v.
a = 𝚺F / m so v2 / r = (-GMm/rs2) / m
v = Δ𝜃 / Δt. or 2𝛑rs / T where T is the period of the satellites orbit.
v2 / r = (-GMm/rs2) / m
(4𝛑2rs2 / T2) / r = -GM/r or
T =
sqrt((4pi^2r_s^3)/(GM_p)