Sect. 6.2: Newton’s Laws + Non-Uniform Circular Motion

1. Sect. 6.2: Newton’s Laws + Non-Uniform Circular Motion

2. Tangential & Radial Acceleration (From Ch. 4) • Consider an object moving in a curved path. If it’s speed |v| is changing,There is an acceleration in the direction of motion≡ Tangential Acceleration at ≡ |dv/dt| • But, There still is also always a radial (centripetal) acceleration perpendicular to the direction of motion. ≡ Radial (Centripetal) Acceleration ar = |ac|≡ (v2/r)

3. Total Acceleration • In general, there are two vector components of the acceleration: • Tangential:at = |dv/dt| • Radial:ar = (v2/r) • Total Acceleration is the vector sum: a = ar+ at • So, in general, there also always two vector components of the net force. That is, ∑F = ma has 2 vector components. • Tangential:∑Ft = mat&Radial:∑Fr = mar = m(v2/r) • The Total Force is the vector sum: ∑F = ∑Ft + ∑Fr = m(ar+ at)

4. Example 6.6 A sphere, mass m, is attached to a cord, length R. It is twirled in a vertical circle about a fixed point O. Find a general expression for the tension T at any instant when the sphere’s speed is v & the angle the cord makes with the vertical is θ. Forces acting are gravity, Fg = mg, & cord tension, T. Solution Resolve Fginto components tangent to the path, Fgt = mg sinθ, & perpendicular to path (radial), Fgr= mg cosθ. 2. Apply Newton’s 2nd Law separately in the two directions: ∑Ft = mg sinθ = mat, so at, = mg sinθ ; ∑Fr = T - mg cosθ = mac = m(v2/R)so T = mg [v2/(Rg) + cosθ]The general expression needed! 3. Evaluate T at top (θ = 180°, cosθ = -1) & at bottom:(θ = 0°, cosθ = 1) Ttop = mg [(vtop)2/(Rg) - 1]; Tbot = mg [(vbot)2/(Rg) + 1]

5. Sect. 6.3: Motion in Accelerated Frames • Recall,Ch. 5:Newton’s Laws technically apply only in inertial (non-accelerated) reference frames.Consider a train moving with respect to the ground at constant velocityvtg to the right. A person is walking in the train at a constant velocity vptwith respect to the train. Newton’s Laws hold for the person walking in the train as long as the train velocity vtg with respect to the ground is constant (no acceleration). But, if the train accelerates with respect to the ground (vtg≠ constant) & the person does an experiment, the result will appear to violate Newton’s Laws because objects will undergo an acceleration in the opposite direction as the train’s acceleration. This acceleration will (seem to be) unexplained because they occur in the absence of forces! vpt vtg

6. More Discussion • Inertial Reference Frame: • Any frame in which Newton’s Laws are valid! • Any reference frame moving with uniform (non-accelerated) motion with respect to an “absolute” frame “fixed” with respect to the stars. • By definition, Newton’s Laws are only valid in inertial frames! ∑F = ma: Not valid in a non-inertial frame! • By definition, a “Force”, as defined in Ch. 5 (& by Newton), is that it comes about from an interaction between objects.

7. Fictitious “Forces” • Suppose a person on a train places a flat object on a flat, horizontal, frictionless surface. There are no horizontal forces on the object. The train accelerates forward. The person will see the flat object accelerate horizontally backward at the same time. Thisappears to violate Newton’s Lawsbecause there is no horizontal force on the object (accelerations are caused by forces). This only appears to violate Newton’s Laws, but actually doesn’t, because the accelerating train isNOT AN Inertial Reference Frame.Based on this experiment, the person will say that there is aFictitious (or Apparent) Forceon the object.

8. Fictitious (Apparent) Forces: • If we insistthat Newton’s 2nd Law holds in a non-inertial reference frame. To make the force equations look as if the reference frame is an inertial one, it’s necessary to introduce Fictitious Forces. • Technically, it’s the coordinate transformation from the inertial frame to the non-inertial one introduces terms on the “ma” side of ∑F = mainertial. If we want eqtns in the non-inertial frame to look like Newton’s Laws, these terms moved to the “F” side & we have ∑“F” = manoninertial. where “F” = F + terms from coord transformation  “Fictitious (Apparent) Forces”.

9. Simple Example of a Fictitious Force You ride in a car going around a curve at constant  constant velocity v.  There is a centripetal acceleration on the car:ac= (v2/r), towards the center of the curve(r = curve radius). Car takes the curve fast enough that, in the passenger seat, you slide to the right & bump against the door. The force between door & your body keeps you from being thrown out of the car. Your reference frame, the car, is a NON-INERTIAL frame.You feel an apparentforce moving you towards the door (away from the center of the curve). A popular & WRONG explanation is that you feel a “Centrifugal Force” away from the center of the curve. “Centrifugal Force” ” is an example of a fictitious force.

10. Simple Example Continued From your viewpoint, in the non-inertial reference frame of the car, there is an apparent force, pointed away from the center of the curve & pushing you to the door. Correct explanation:Before car starts to round the curve, it (& you) are going in a straight line at constant velocity. As car enters the curve, from N’s 1st Law, you will first tend to keep moving straight. In the inertial frame of the Earth, You feel a real force (friction), acting inward towards the curve center between you & the car seat that makes you stay in your seat. N’s Lawssay that you will tend to continue to move in your original straight line until you bump against the door.

11. Example 6.7