1. Chapter 6 Circular Motion and Other Applications of Newton’s Laws EXAMPLES

2. Example 6.1 Conical Pendulum Find an expression for v? The object is in equilibrium in the vertical direction and undergoes UCM in the horizontal direction Components of T T x = Tsinθ T y = Tcosθ T x = ma C = mv 2 /r & T y = mg  T x = Tsinθ = mv 2 /r (1) T y = Tcosθ = mg (2) Dividing (1) by (2): Sinθ/cosθ = tanθ = v 2 /rg Since: r = Lsinθ  tanθ = v 2 /Lsinθg Solving for v: v is independent of m v TyTy TxTx θ

3. Example 6.2 Force on Revolving Ball (Similar to Example 6.2 Text Book) Find the tension T ? If: m = 0.150-kg, r = 0.600 m, f = 2 rev/s (T = 0.500s) Assumption: Circular path is  in horizontal plane, so: θ  0  cos(θ)  1 Newton’s 2 nd Law: ∑F = ma  T = ma x = ma C = mv 2 /r But: v = 2πr/T = 7.54 m/s  T = (0.150kg)(7.54m/s) 2 /0.600m)  T = 14.2N (Tension) v T

4. Example 6.3 Highway Curves A 1000 kg car rounds a curve on a flat road of radius 50.0m at a speed of 50 km/hr (14m/s). Find the Coefficient of Static Friction (  s ) if the car follows the curve and make the turn successfully.

5. Example 6.3 Highway Curves, cont Radial (x): ΣF x = ma c Car stays in the curve if and only if ƒ s,max ≥ ma c  ƒ s,max = ma c = mv 2 /r Vertical (y): n = mg Since: ƒ s,max =  s n =  s mg   s mg= mv 2 /r (cancel m) Solving for  s :

6. Example 6.4 Banked Curve These are designed with friction equaling zero A car is traveling with speed v around a curve of radius r, determine a formula for the angle (θ) at which a road should be banked so that no friction is required. Solution: There is a component of the normal force (nsinθ) that supplies the centripetal force Car is moving along a horizontal circle so a c is horizontal too. θ

7. Example 6.4 Banked Curve, cont Components of n: n x = nsinθ n y = ncosθ x direction: ΣF x =ma x  n x = ma C = mv 2 /r  nsinθ = mv 2 /r (1) y direction: ΣF y = 0  ncosθ = mg (2) Dividing (1) by (2): Sinθ/cosθ = tanθ = v 2 /rg (Similar to The Conical Pendulum)  θ n y = ncosθ n x = nsinθ

8. Example 6.5 Force on Revolving Ball. Top and Bottom of Circle (Vertical) Similar to Examples 6.5 (Text Book) Just change T by n. Given: R, v, m. Use: ΣF =ma c = mv 2 /R, to find tensions T top & T bot The tension at the bottom is maximum T bot – mg = mv 2 /R  The tension at the top is minimum T top + mg = mv 2 /r  If T top = 0, then v 2 /r = g  acac acac

9. Example 6.6 Vertical Circle with Non- Uniform Speed The gravitational force exerts a tangential force on the object Look at the components of F g The tension at any point can be found: T – mgcosθ = ma c = mv 2 /R  T = mv 2 /R + mgcosθ 

10. Material from the book to Study!!! Objective Questions: 1-4 Conceptual Questions: 2-3-8 Problems: 2-7-9-11-18-54-57 Material for the Final Exam