Kinematics, Forces and Differential Eqns Definitions of velocity and acceleration: v = dtdx x = vdta = 2 2 dtx ddtdv v = adtDifferential Equations and Kinematics/Force s: F = m dtdv Insert forces in left hand side, then separate variables, then integrate F = m 2 2 dtx dInsert forces into left hand side and compare to0 2 2 2 x dtx dNote: =2 /T Ex: The force on a car of mass m is given by F(t)=4t+2, find an expression for the velocity as a function of time v(t) if the car starts from rest.dtdv m ttF2 4 ) ( (set the forces in a given direction equal to dtdv m ) dv dtm t) 2 4 ( (separate variables to get all t on one side and all v on the other) dv dtm tv t0 0 ) 2 4 ( (integrate both sides and choose appropriate limits of integration) m tm ttv 2 2 4 ) ( 2 (do integral and use correct limits) Ex: Air resistance.Assume the force of air resistance on a falling object is given by F = -kv. Determine the velocity as a function of time for a dropped object. Fdtdv m mgkv (treat downward as positive since object is falling) mdv dtmgkv ) ( (make the whole force side one term to separate!) ) ( mgkv dv m dtso v tmgkv dv m dt0 0 ) ( (separate variables, and integrate) mgmgkv kmgkmgkv km t) ( ln 1 ) ln( 1 ) ln( 1 (now just use algebra to solve for v) mgkv e mgm tk) ( and solving for v you get… ] 1 [ ) ( m tke kmgtv This expression for v(t) mankes sense since at t=0, v=0. And for t= inifinity. v=mg/k which is terminal velocity.
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Fe is electric force [Newtons]+ charges feel force in same dir of the electric field (E), - charges feel force in opp. dir of electric field (E)
E is electric field [N/Coulomb or Volts/meter]+ charges create electric fields that point radially outward from charge, - charges create E pointing inward
Uelectric is electric potential energy [Joules]Electric Poetential Energy (U) is another form of energy that objects can have
V is electric potential [J/Coulomb or Volts]Electric Potential (V) at a point is the Electric Potential Energy (U) 1C of charge would have at that point
Note: + and - charges both feel a force toward lower PEelectric , also Electric fields E point toward lower V
Solving 2D electric force (vector) problems1. Draw the forces exerted on the charge you are concerned with(These are forces ON charge Q, not the forces charge Q is exerting on other charges)
2. Find the size of each force using F=kq1q2/r2
3. Break forces into vertical and horizontal components Fy and Fx (For completely vertical or horizontal forces one component will be zero, the other is + F)
(For diagonal forces you need to use sin and cos to break the vector into components)
(Up or Right is a positive component, Left or Down is a negative component)
4. Add up all the Fx and Fy to get the components of the total Force vector Fxtot and
Fytot
i.e. for the example shown Fxtot = Fa cos + 0 + (-Fc) Fytot = Fa sin + (-Fb) + 0
5. Find magnitude of total Force using the Pythagorean Theorem i.e. Ftot2 = Fxtot
2
+ Fytot2
6. Find the angle from horizontal by using = tan-1
(Fytot/Fxtot) (Note: This angle is always the angle from the “nearest” x-axis to the total Force vector)
Differential equations and R-L-C circuitsWhen you are asked to find something (current, voltage, charge, etc.) as a function of time, use the loop rule
to write and solve a differential equation.
RC circuit
-for the voltage drop across a resistor use – I R = -(dQ/dt)R
-for the voltage drop across a capacitor use –Q/C
-for most charging RC circuits you can use
0C
Q R
dt
dQ
And you separate Q and t, then integrate to get,
)1()( RC
t
o eQt Q which means that RC
t
o RC
t
o e I e RC Q
dt dQt I )(
Note: = RC gives you an idea of how long it takes to charge/discharge
-for most discharging RC circuits you can use
0C
Q R
dt
dQ Note: You often keep the dQ/dt term negative for a discharging
capacitor since the Q flows through R= Q o –Q still on capacitor
And you separate Q and t and integrate to get,
RC
t
oeQt Q )( which means that RC
t
o RC
t
o e I e RC
Q
dt
dQt I )(
Note: Q o = CVo for charging and discharging capacitors (Vo is the maximum voltage across capacitor)
FB = qvBsin vB (q is charge, v is speed, B is magnetic field, is angle between v and B)- The direction of force on + charge is given by the Right hand rule (Very-Bad-Finger)
- If the charge is negative the force is in the opposite direction
- Magnetic forces never do Work (since FB is always perpendicular to motion W=Fdcos90=0)
- Magnetic forces often make charges (q) of mass m travel in circles of radius r given by,
r = mv/qBNote: If you want a charged particle to travel in a straight line (“velocity selector”), create an electric field E so
that the forces cancel, i.e. speed is ratio of E to B
v=E/B (since FB = FE or qvB = qE)Note: The forces have to be of equal size, not the fields! (i.e. FB = FE , but E does not equal B)
Magnetic force on wire
FB = ILB sin LB -to find direction of FB use the same right hand rule (except v is now direction of I)
Magnetic fields
Biot Savart Law 2
ˆ
4 r
r l d I B
o
(true for all cases, but rarely used)
Long straight wire B = oI/2 r [Tesla] o = x 10-7
T m/A
- The magnetic field from a long straight wire is directed along a circle centered at wire with direction given by
right hand rule (Thumb in direction of current, fingers curl in direction of B)
- Note: Wires with I in same direction will attract, Wires with I in opp. Direction will repel