1. Chapter Opener. Caption: Newton’s laws are fundamental in physics
Chapter Opener. Caption: Newton’s laws are fundamental in physics. These photos show two situations of using Newton’s laws which involve some new elements in addition to those discussed in the previous Chapter. The downhill skier illustrates friction on an incline, although at this moment she is not touching the snow, and so is retarded only by air resistance which is a velocity-dependent force (an optional topic in this Chapter). The people on the rotating amusement park ride below illustrate the dynamics of circular motion.
Physics”

2. Example 5-11 (Estimate)
Estimate the force a person must exert on a string attached to a 0.15-kg ball to
make the ball revolve in a horizontal circle of radius 0.6 m. The ball makes 2
rev/s. Ignore the string mass. m = 0.15 kg, r = 0.6 m, f = 2 rev/s  T = 0.5 s
Assumption: Circular path is  in a horizontal plane, so φ  0  cos(φ)  1
Figure 5-17.
Answer: Ignoring the weight of the ball, so FT is essentially horizontal, we find FT = 14 N.

3. Newton’s 2nd Law: ∑F = ma  FTx = max= maR = m(v2/r)
Example 5-11 (Estimate)
Estimate the force a person must exert on a string attached to a 0.15-kg ball to
make the ball revolve in a horizontal circle of radius 0.6 m. The ball makes 2
rev/s. Ignore the string mass. m = 0.15 kg, r = 0.6 m, f = 2 rev/s  T = 0.5 s
Assumption: Circular path is  in a horizontal plane, so φ  0  cos(φ)  1
Newton’s 2nd Law: ∑F = ma  FTx = max= maR = m(v2/r)
v = (2πr/T) = 7.54 m/s
So, the tension is (approximately) FTx  14 N
Figure 5-17.
Answer: Ignoring the weight of the ball, so FT is essentially horizontal, we find FT = 14 N.

4. Example 5-12: Revolving ball (vertical circle)
A ball, mass m = 0.1 kg on the end of a (massless) cord of length r = 1.1 m cord is swung in a vertical circle. Calculate:
a. The minimum speed the ball must have at the top of its arc so that the ball continues moving in a circle.
b. The tension in the cord at the bottom of the arc, assuming that there the ball is moving at twice the speed found in part a.
Hint: The minimum speed at the top will happen for the minimum tension FT1
Figure Caption: Example 5–12. Freebody diagrams for positions 1 and 2.
Solution: See the freebody diagrams.
The minimum speed occurs when the tension is zero at the top; v = m/s.
Substitute to find FT2 = 7.35 N.

5. A Similar Problem ∑F = maR Newton’s 2nd Law:
r = 0.72 m, v = 4 m/s = constant
m = 0.3 kg, FT1 = ?, FT2 = ?
Newton’s 2nd Law:
∑F = maR

6. A Similar Problem ∑F = maR Newton’s 2nd Law:
r = 0.72 m, v = 4 m/s = constant
m = 0.3 kg, FT1 = ?, FT2 = ?
Newton’s 2nd Law:
∑F = maR
At the top of the circle, vertical forces:
(down is positive!)
FT1 + mg = m(v2/r)  FT1 = 3.73 N
At the bottom of the circle, vertical forces:
(up is positive!)
FT2 - mg = m(v2/r)  FT2 = 9.61 N

7. Example: F, p. 124
A Ferris wheel rider moves in a vertical circle of radius r at constant speed v. Is the normal force FN1 that the seat exerts on the rider at the top of the wheel
a. less than, b. more than, or c. equal to the normal force FN2 that the seat exerts on the rider at the bottom of the wheel?
Use
Newton’s 2nd Law: ∑F = maR at top & bottom, solve for normal force & compare. 1 2

8. Conceptual Example Newton’s 2nd Law: ∑F = ma  FTx = maR = m(v2/r)
A tether ball is hit so that it revolves
around a pole in a circle of radius r at
constant speed v. In what direction is the
acceleration & what force causes it?
Newton’s 2nd Law: ∑F = ma
x: ∑Fx = max
 FTx = maR = m(v2/r)
y: ∑Fy = may = 0
 FTy - mg = 0, FTy = mg

9. Example 5-13: Conical pendulum
A ball of mass m, suspended by a cord of length ℓ, revolves in a circle of radius r = ℓsinθ, where θ is the angle the string makes with the vertical.
a. In what direction is the ball’s acceleration & what causes it?
b. Calculate the speed & period (time for one revolution) of the ball in terms of ℓ, θ, g, & m.
Figure Caption: Example 5–13. Conical pendulum.
Answer: a. The acceleration is towards the center of the circle, and it comes from the horizontal component of the tension in the cord.
b. See text.

10. Sect. 5-4: Highway Curves: Banked & Unbanked
Case 1: Unbanked Curve. When a car goes around a curve, there must be a net force toward the center of the circle of which the curve is an arc. For flat road, that force is supplied by static friction.
“Centripetal Force”
No static friction?
 No Centripetal Force?
 The car goes straight!
Never a “Centrifugal Force”!!!

11. Example 5-14: Skidding on a curve
A car, mass m = 1,000 kg rounds a curve on a flat road of radius r = 50 m at a constant speed v =15 m/s (54 km/h). Will the car follow the curve, or will it skid? Assume:
a. The pavement is dry & the coefficient of static friction is μs = 0.6.
b. The pavement is icy & μs = 0.25.
Free Body
Diagram
Figure Caption: Example 5–14. Forces on a car rounding a curve on a flat road. (a) Front view, (b) top view.
Solution: The normal force equals the weight, and the centripetal force is provided by the frictional force (if sufficient). The required centripetal force is 4500 N.
The maximum frictional force is 5880 N, so the car follows the curve.
The maximum frictional force is 2450 N, so the car will skid.

12. Example 5-14: Skidding on a curve
A car, mass m = 1,000 kg rounds a curve on a flat road of radius r = 50 m at a constant speed v =15 m/s (54 km/h). Will the car follow the curve, or will it skid? Assume:
a. The pavement is dry & the coefficient of static friction is μs = 0.6.
b. The pavement is icy & μs = 0.25.
Newton’s 2nd Law: ∑F = ma
x: ∑Fx = max  Ffr = maR = m(v2/r)
y: ∑Fy = may = 0  FN - mg = 0; FN = mg
Maximum static friction: Ffr = μsFN
Free Body
Diagram
Figure Caption: Example 5–14. Forces on a car rounding a curve on a flat road. (a) Front view, (b) top view.
Solution: The normal force equals the weight, and the centripetal force is provided by the frictional force (if sufficient). The required centripetal force is 4500 N.
The maximum frictional force is 5880 N, so the car follows the curve.
The maximum frictional force is 2450 N, so the car will skid.

13. The kinetic friction force is smaller than the static one.
If the friction force is insufficient, the car will tend to move more nearly in a straight line, as the skid marks show.
As long as the tires don’t slip, the friction is static. If the tires do start to slip, the friction is kinetic, which is bad in 2 ways:
The kinetic friction force is smaller than the static one.
The static friction force can point toward the center of the circle, but the kinetic friction force opposes the direction of motion, making it very difficult to regain control of the car & continue around the curve.
Figure Caption: Race car heading into a curve. From the tire marks we see that most cars experienced a sufficient friction force to give them the needed centripetal acceleration for rounding the curve safely. But, we also see tire tracks of cars on which there was not sufficient force—and which unfortunately followed more nearly straight-line paths.

14. Banked Curves Newton’s 2nd Law x: ∑Fx = max  FNx = m(v2/r)
Banking a curve helps keep cars from skidding. For every banked curve, there is one speed v at which the entire “centripetal force” is supplied by the horizontal component of the normal force FN & no friction is required.
Newton’s 2nd Law
tells us what speed v this is:
x: ∑Fx = max  FNx = m(v2/r)
or (solve for v) Aa
Also, y: ∑Fy = may = 0  FN cosθ - mg = 0
or FNcosθ = mg (solve for FN)
Figure Caption: Normal force on a car rounding a banked curve, resolved into its horizontal and vertical components. The centripetal acceleration is horizontal (not parallel to the sloping road). The friction force on the tires, not shown, could point up or down along the slope, depending on the car’s speed. The friction force will be zero for one particular speed.

15. Example 5-15: Banking angle
a. For a car traveling with speed v around a curve of radius r, find a formula for the angle θ at which a road should be banked so that no friction is required.
b. Calculate this angle for an expressway off-ramp curve of radius r = 50 m at a design speed of v = 14 m/s (50 km/h)?
Answer: a. Set FN = mg in previous equation. Find tan θ = v2/rg.
b. Tan θ = 0.40, so θ = 22°.

16. Example 5-15: Banking angle
a. For a car traveling with speed v around a curve of radius r, find a formula for the angle θ at which a road should be banked so that no friction is required.
b. Calculate this angle for an expressway off-ramp curve of radius r = 50 m at a design speed of v = 14 m/s (50 km/h)?
Answer: a. Set FN = mg in previous equation. Find tan θ = v2/rg.
b. Tan θ = 0.40, so θ = 22°.
Newton’s 2nd Law
x: ∑Fx = max  FNx = m(v2/r) or FNsinθ = m(v2/r) (1)
y: ∑Fy = may = 0  FN cosθ - mg = 0 or FNcosθ = mg (2)
Dividing (2) by (1) gives: tanθ = [(v2)/(rg)]
Putting in the given numbers, tanθ = 0.4 or θ = 22º