; coache notes ap particle
Learning Center
Plans & pricing Sign in
Sign Out
Your Federal Quarterly Tax Payments are due April 15th Get Help Now >>

coache notes ap particle


  • pg 1
									                  Forces due to Friction
  Friction: A force between the contacted surfaces
  of two objects that resists motion.
If an object is not moving, that does not mean that
there are no friction forces at work:

                          In this case, there is no
                          motion, nor any additional
                          external force that may
                          produce motion, so there
                mg        is not friction present!
No Motion F = fstatic
    F            fs
                           Accelerated Motion
                           F            fk
 Uniform Motion

F                     fk
                           If Fmax > fs-max then acc.

                                  and then:
        F = fk                    F > fkinetic
                                max fs

Friction (N)      fs

                    time (s)
Static friction builds to a maximum value at which
point the object “breaks free” -- it accelerates and
kinetic friction is present!
        No Acceleration  Static Friction
The ratio of the maximum force of static friction
to the magnitude of the normal force is called the
coefficient of static friction:
                                  s = f s

The ratio of the magnitude of the force of kinetic
friction to the magnitude of the normal force is
called the coefficient of kinetic friction:
                                   k = fk
Examine the right side:

    fs                    ∑Fx = 0 = F – N
                 F         ∑Fy = 0 = fs – .5mg

                          µs = fs
         .5mg        N = F = fs       = .5mg
= .5(75)(9.80)               µs          µs
     .41             = 900 N
Will the blocks accelerate, and if so, what will
that acceleration be?



Ø = 34˚                 µs = .24
m1 = 9.5 kg
                       µk = .15
m2 = 2.6 kg
                 T1           ∑Fy2 = 0 = m2g - T2

m1gsinø                     m 2g

            ø    m1gcosø       T1 = T2 = T
                     If the block will stay, then:
                  ∑Fx1 = 0 = T - (fs + m1gsinø)
                   ∑Fy1 = 0 = N - m1gcosø
                    N = m1gcosø
T = fs + m1gsinø
m2g = µs(m1gcosø) + m1gsinø
m2 = m1(µscosø + sinø)

    2.6 < 8.5 therefore: T < fs + m1gsinø
 The tension in the string is not strong enough!
OR: ∑Fx = 0 = T + fs - m1gsinø
T = m1gsinø - fs
m2g = m1gsinø - µsm1gcosø

 2.6 < 7.2 again, T is not enough to hold
To find the acceleration:
∑Fx = m1a1 = T + fk - m1gsinø
∑Fy = 0 = N - m1gcosø

 ∑Fy = m2a2 = T - m2g
in terms of the magnitude of a: a1 = a2 = a
  T = m1a + m1gsinø - fk
 T = m2a + m2g
m1a + m1gsinø - fk = m2a + m2g
m1a + m1gsinø - µkm1gcosø = m2a + m2g
a(m1 + m2) = m2g + m1g(µkcosø - sinø)

a = [m2g + m1g(µkcosø - sinø)]
             (m1 + m2)

a = 25 + 93[(.15)(.83) - (.56)]     = - 1.3 m/s2
Note well what the negative sign means:
 • for block 1, acceleration is in the negative x
 direction- opposite of up (+) the incline.
 • for block 2, opposite of g (which we used as +)
    The dynamics of Uniform Circular Motion
Remember that for any object traveling in
uniform circular motion, the magnitude of the
net (centripetal) force acting upon the object is:
             ∑F = ma = mv2/r
For the purposes of discussing motion of an
object in uniform circular motion, we will not
use x and y axes!
Instead, vector will be described in the z direction
(perpendicular to the plane of motion) and r
direction (along the radius of the path)!
We will consider 3 specific examples of forces
that act centripetally:
             The Conical Pendulum
      L     ø            r


    m                            mg
∑Fr = ma = mv2/R = Tsinø
∑Fz = 0 = Tcosø - mg
Setting the components in ratio form:
     Tsinø = mv2/R
     Tcosø        mg

 Solving for v:     v = √ Rgtanø

 This equation gives the constant speed needed
 for the pendulum to maintain its circle.
The constant speed of a body would be:
        v = 2πR
Substituting and solving for time:

    T = 2π (Lcosø)/g

This would give the period of the motion, which
is independent of the mass of the bob!
The Rotor


                    The Rotor

An object is kept in uniform circular motion by
frictional forces!
      ∑Fz = 0 = fs - mg       fs = mg
       ∑Fr = mar = N         N = mv2/R

Solving these equations together we can derive
an equation for speed needed to prevent
                    v = √ gR/µs
The Banked Curve
         The Banked Curve

             ∑Fz = 0 = Ncosø - mg

N              ∑Fr = mar = Nsinø
                  manipulating these
    ø             equations will give:
                      tanø = v2/Rg

        mg       this is the angle-speed
               combination where friction
                      is not present!
   tanø = v2/Rg         NOTE:

• the angle of the bank is independent of the mass
of the vehicle or even friction!
 • when a banked curve is constructed, the road is
 banked according to an average calculated speed.
• as with the conical pendulum, the constant speed
needed to maintain a given banked curve
(neglecting any frictional forces):

                   v = Rgtanø
                Drag Forces and Projectiles
Due to air resistance, objects will only accelerate
until reaching a final /maximum speed.
   That max. speed is called terminal velocity.
         Terminal velocity depends on the
         properties of the falling object (size and
         shape and density) as well as the properties
         of the fluid (esp. density).

   One particular characteristic of drag forces is
   that they depend on velocity.
        Air Resistance (Aerodynamic Drag)
Unlike kinetic friction, air resistance (Drag) is not
constant but increases as speed increases.
The amount of Drag will depend on the shape of
the object (drag coefficient [C]), the density of the
air (ρ), the cross sectional area (A) and the square
of the velocity of the object:

                   D = .5CρAv2
Terminal Velocity occurs when the object stops
accelerating and the Drag equals weight!
∑Fy = 0 = mg - D

                   mg = .5CρAv2
For a relatively tiny particle traveling through a
thick fluid, Drag is proportional to only velocity:

                      D = bv

b is a constant value dependent upon the
characteristics of the object and the fluid

Terminal velocity in the fluid would still occur
when the Drag equals the weight force:

                  bvT = mg

To top