Physics”

Example 5-11 (Estimate)

Estimate the force a person must exert on a string attached to a 0.15-kg ball tomake the ball revolve in a horizontal circle of radius 0.6 m. The ball makes 2rev/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

Example 5-11 (Estimate)

Estimate the force a person must exert on a string attached to a 0.15-kg ball tomake the ball revolve in a horizontal circle of radius 0.6 m. The ball makes 2rev/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

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

A Similar Problemr = 0.72 m, v = 4 m/s = constantm = 0.3 kg, FT1 = ?, FT2 = ?

Newton’s 2nd Law: ∑F = maR

A Similar Problemr = 0.72 m, v = 4 m/s = constantm = 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

Example: F, p. 124A 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? UseNewton’s 2nd Law: ∑F = maR

at top & bottom, solve for normal force & compare.

1

2

Conceptual ExampleA tether ball is hit so that it revolvesaround a pole in a circle of radius r atconstant speed v. In what direction is theacceleration & what force causes it?

Newton’s 2nd Law: ∑F = max: ∑Fx = max

FTx = maR = m(v2/r)

y: ∑Fy = may = 0 FTy - mg = 0, FTy = mg

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.

Sect. 5-4: Highway Curves: Banked & UnbankedCase 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”!!!

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 BodyDiagram

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 BodyDiagram

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:

1. The kinetic friction force is smaller than the static one.

2. 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.

Banked CurvesBanking 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 Lawtells 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)

Example 5-15: Banking anglea. 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)?

Example 5-15: Banking anglea. 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)?

Newton’s 2nd Lawx: ∑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