Lecture 3: Forces and Dynamics

Dynamics

Dynamics
  • Deals with the study of forces and their effect on motion.
Newton's Second Law
  • F=maF = ma is the fundamental equation.
    • F represents force.
    • m represents mass.
    • a represents acceleration.
  • Applies to a point mass or an object that can be treated as a point mass.
Non-inertial Reference Frames
  • Lab reference frame is considered a non-inertial reference frame.
  • In a non-inertial reference frame, the equation Fnet=maF_{net} = ma does not hold true directly. A fictitious force needs to be introduced.
  • Fictitious Force: Ffictitious=mAF_{fictitious} = -mA, where A is the acceleration of the non-inertial reference frame.
  • The transformation of acceleration from an inertial frame to a non-inertial frame is given by: a=a+Aa = a' + A, where:
    • aa is the acceleration in the inertial frame.
    • aa' is the acceleration in the non-inertial frame.
    • AA is the acceleration of the non-inertial frame with respect to the inertial frame.
  • Therefore, a=a+(A)=aAa' = a + (-A) = a - A
  • The net force in the inertial frame is Fnet=m(a+A)F_{net} = m(a'+ A)
  • In the non-inertial frame, the modified Newton's second law is:
    • Fnet+(mA)=maF_{net} + (-mA) = ma'
    • F<em>net+F</em>fictitious=maF<em>{net} + F</em>{fictitious} = ma'
    • If one were to use the non-intertial reference frame, then Fnet=maF_{net} = ma'
Example: Pendulum in an Accelerating Car
  • Consider a pendulum inside a car accelerating with acceleration AA.
  • The tension in the string is T.
  • In the non-inertial frame (inside the car), the forces acting on the pendulum bob are:
    • Tension T.
    • Weight mg.
    • Fictitious force mA-mA.
  • The pendulum will deflect at an angle such that the net force balances.
  • Tx=mAT_x = mA
  • Ty=mgT_y = mg
  • The time period of the pendulum is T=2πlg<em>effT = 2\pi \sqrt{\frac{l}{g<em>{eff}}}, where g</em>effg</em>{eff} is the effective acceleration due to gravity.
  • Here, geff=g2+A2g_{eff} = \sqrt{g^2 + A^2}
Steps to Solve F=maF = ma Problems
  • Applicable to n-object systems.
  • Write down F=maF = ma for each object. This will give you n equations.
  • Identify all unknown variables (n+m in total).
  • Identify all forces acting on each object.
  • Identify constraints on the motion of the objects. These constraints will give you m additional equations.
  • Solve the system of equations to find the unknowns.
Example: Two Blocks Connected by a String
  • Two blocks with masses M and m are connected by a string over a pulley.
  • M is on a horizontal surface, and m is hanging vertically.
  • Forces on M: Tension T and friction.
  • Forces on m: Tension T and weight mg.
  • Equations for M:
    • F<em>x=T=Ma</em>x\sum F<em>x = T = Ma</em>x
    • Fy=NMg=0\sum F_y = N - Mg = 0
  • Equations for m:
    • F<em>y=mgT=ma</em>y\sum F<em>y = mg - T = ma</em>y
  • Constraint: a<em>x=a</em>y=aa<em>x = a</em>y = a
  • Solve for T and a.
Example: Block on an Inclined Plane in an Accelerating System
  • A block of mass m is on an inclined plane (angle θ) inside an accelerating system (acceleration A<em>xA<em>x). The acceleration in y direction is zero (A</em>y=0A</em>y = 0).
  • The acceleration in the non-inertial frame is a=a+Aa = a' + A
  • a<em>x=a</em>x+(Ax)a<em>x = a'</em>x + (-A_x)
  • a<em>y=a</em>y+0a<em>y = a'</em>y + 0
  • Forces: Normal force N, weight mg.
  • Equations:
    • Nsin(θ)=maxN \sin(\theta) = m a_x (∑Fx = max)
    • Ncos(θ)=mgN \cos(\theta) = mg (∑Fy = may)
  • Constraint Example: Two masses connected by a string over a pulley.
  • m1 and m2 are connected by the string the constraints that relate the acceleration is a1=a2a1 = -a2
  • m<em>1a</em>1=Tm1gm<em>1 a</em>1 = T - m_1g
  • m<em>2a</em>2=m2gTm<em>2 a</em>2 = m_2g -T
  • d2x<em>1dt2=d2x</em>2dt2\frac{d^2x<em>1}{dt^2} = - \frac{d^2x</em>2}{dt^2}