Extension of Circular Motion Newton’s Laws of Circular Motion Newton’s ... Kings High School . Review from Chapter 4 SUniform Circular Motion ... 14, 15, 16, 18/ (Section

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  • S

    Extension of Circular Motion & Newtons Laws

    Chapter 6

    Mrs. Warren

    Kings High School

  • Review from Chapter 4

    S Uniform Circular Motion

    S Centripetal Acceleration

  • Uniform Circular Motion,

    Force

    A force is associated with the

    centripetal acceleration.

    The force is also directed

    toward the center of the circle.

    Applying Newtons Second Law along the radial direction

    gives

    rF

    2

    c

    vF ma m

    r= =

    Section 6.1

  • Uniform Circular Motion, cont.

    SA force causing a centripetal

    acceleration acts toward the center of

    the circle.

    SIt causes a change in the direction of

    the velocity vector.

    SIf the force vanishes, the object

    would move in a straight-line path

    tangent to the circle.

    S See various release points in

    the active figure

    Section 6.1

  • S A small ball of mass, m, is suspended

    from a string of length, L. The ball

    revolves with constant speed, v, in a

    horizontal circle of radius, r. Find an

    expression for v in terms of the geometry.

    The Conical

    Pendulum

    sin tanv Lg q q=

  • The Conical Pendulum

    S A puck of mass 0.500 kg is attached to the end of a cord. The puck moves in a horizontal

    circle of radius 1.50 m. If the cord can withstand a maximum tension of 50.0 N, what is

    the maximum speed at which the puck can move before the cord breaks? Assume the

    string remains horizontal during the motion.

    Trv

    m=

  • The Flat

    Curve Model the car as a particle in

    uniform circular motion in the horizontal direction.

    Model the car as a particle in equilibrium in the vertical direction.

    The force of static friction supplies the centripetal force.

    The maximum speed at which the car can negotiate the curve is:

    Note, this does not depend on the mass of the car.

    sv grm=

  • The Flat Curve

    S A 1500 kg car moving on a flat, horizontal road negotiates a curve as shown on the

    previous slide. If the radius of the curve is 35.0 m and the coefficient of static friction

    between the tires and dry pavement is 0.523, find the maximum speed the car can

    have and still make the turn successfully.

  • The Banked Roadway

    SThese are designed with friction equaling

    zero.

    SModel the car as a particle in equilibrium in

    the vertical direction.

    SModel the car as a particle in uniform circular

    motion in the horizontal direction.

    SThere is a component of the normal force that

    supplies the centripetal force.

    SThe angle of bank is found from

  • The Banked Curve

    S A civil engineer wishes to redesign the curved roadway in such a way that a car will not

    have to rely on friction to round the curve without skidding. In other words, a car moving

    at the designated speed can negotiate the curve even when the road is covered with ice.

    Such a road is usually banked, which means that the roadway is tilted toward the inside

    of the curve. Suppose the designated speed for the road is to be 30.0 mi/h and the radius

    of the curve is 35.0 m. At what angle should the curve be banked?

  • Ferris Wheel S A child of mass m rides

    on a Ferris wheel as

    shown. The child moves

    in a vertical circle of

    radius 10.0 m at a

    constant speed of 3.00

    m/s. Determine the force

    exerted by the seat on the

    child at the bottom of the

    ride. Express your answer

    in terms of the weight of

    the child, mg. Do the

    same when the child is at

    the top of the ride.

  • Non-Uniform Circular Motion

    SThe acceleration and force have tangential

    components.

    S produces the centripetal acceleration

    S produces the tangential acceleration

    SThe total force is

    S r t= + F F F

    tF

    rF

    Section 6.2

  • Vertical Circle with Non-

    Uniform Speed

    SA small sphere of mass, m, is attached

    to the end of a cord of length R and set

    into motion in a vertical circle about a

    fixed point O as illustrated. Determine the

    tangential acceleration of the sphere and

    the tension in the cord at any instant when

    the speed of the sphere is v and the cord

    makes an angle with the vertical.

    2

    cosv

    T mgRg

    q

    = +

    Section 6.2

  • Motion in Accelerated Frames

    SA fictitious force results from an accelerated frame of reference.

    S The fictitious force is due to observations made in an accelerated

    frame.

    S A fictitious force appears to act on an object in the same way as a real

    force, but you cannot identify a second object for the fictitious force.

    S Remember that real forces are always interactions between two objects.

    S Simple fictitious forces appear to act in the direction opposite that of

    the acceleration of the non-inertial frame.

    Section 6.3

  • Centrifugal Force

  • Coriolis Force

  • Motion with Resistive Forces

    SThe medium exerts a resistive force, , on an object moving through the medium.

    SThe magnitude of. .

    SThe direction of is

    SThe magnitude of can depend on the speed in complex ways.

    SWe will discuss only two:

    S is proportional to v

    S Good approximation for

    S is proportional to v2

    S Good approximation for

    R

    R

    R

    Section 6.4

    R

    R

    R

  • Resistive Force Proportional

    To Speed

    SThe resistive force can be expressed as

    Sb depends on the property of the medium, and on the shape

    and dimensions of the object.

    SThe negative sign indicates is in the opposite direction to .

    = -bR v

    R v

    Section 6.4

  • Resistive Force Proportional

    To Speed, Example

    SAssume a small sphere of mass m is

    released from rest in a liquid.

    SForces acting on it are:

    S Resistive force

    S Gravitational force

    SAnalyzing the motion results in

    - = =

    = = -

    dvmg bv ma m

    dt

    dv ba g v

    dt m

    Section 6.4

  • Resistive Force Proportional

    To Speed, Example, cont.

    SInitially, v = 0 and dv/dt = g

    SAs t increases, R increases and a decreases

    SThe acceleration approaches 0 when R mg

    SAt this point, v approaches the terminal speed of the object.

  • Terminal Speed

    STo find the terminal speed, let a = 0

    SSolving the differential equation gives

    St is the time constant and

    t = m/b

    Section 6.4

  • Terminal Speed

    S A small sphere of mass 2.00 g is released from rest in a large vessel filled with

    oil, where it experiences a resistive force proportional to its speed. The sphere

    reaches a terminal speed of 5.00 cm/s. Determine the time constant and the

    time at which the sphere reaches 90.0% of its terminal speed.

  • Resistive Force Proportional

    To v2

    S For objects moving at high speeds through air, the resistive force is approximately equal to the square of the speed.

    S R = DrAv2

    S D is a dimensionless empirical quantity called the drag coefficient.

    S r is the density of air.

    S A is the cross-sectional area of the object.

    S v is the speed of the object.

    Section 6.4

  • Resistive Force Proportional

    To v2, example

    SAnalysis of an object falling

    through air accounting for air

    resistance.

    r

    r

    = - =

    = -

    2

    2

    1

    2

    2

    F mg D Av ma

    D Aa g v

    m

  • Resistive Force Proportional

    To v2, Terminal Speed

    SThe terminal speed will occur

    when the acceleration goes to

    zero.

    SSolving the previous equation

    gives

    r=

    2T

    mgv

    D A

  • Example: Skysurfer

    SStep from plane

    S Initial velocity is 0

    S Gravity causes downward

    acceleration

    S Downward speed increases, but so

    does upward resistive force

    SEventually, downward force of gravity

    equals upward resistive force

    S Traveling at terminal speed

    Section 6.4

  • Skysurfer, cont.

    SOpen parachute

    S Some time after reaching terminal speed, the parachute is

    opened.

    S Produces a drastic increase in the upward resistive force

    S Net force, and acceleration, are now upward

    S The downward velocity decreases.

    S Eventually a new, smaller, terminal speed is reached.

    Section 6.4

  • HW

    S Objective Questions: 1, 4, 5, 6, 7

    S Conceptual Questions: 4, 5, 6, 8

    S Problems: (Section 6.1): 1, 2, 6, 7, 8, 9/ (Section 6.2): 13,

    14, 15, 16, 18/ (Section 6.3): 21, 22/ (Section 6.4): 27, 28,

    30, 31

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