aka…U.C.M.:An object moving in a

perfect circle at a constant speed.

The greater the speed, the faster the object changes direction, the greater the centripetal acceleration.

If comparing two spots on a single rotating object, when placed at a spot of greater radius, the speed is greater, the centripetal acceleration is greater; therefore, the Net force is greater.

Period is the same for both people!

What’s the centripetal acceleration of the bobsled during each section? Speed doesn’t change… assume very little friction. ac = v2/r ac = 35 m/s2 for 33 m radius ac= 48 m/s2 for 24 m radius

As the radius decreases (gets smaller/tighter), the ac increases.

The model airplane of mass 0.90 kg moves at a constant speed on a circle path that is parallel to the ground. The guideline is a length of 17 m. Determine tension FT in the guideline for speeds 19 m/s and 38 m/s.

FT = 19 N FT = 76 N When v is doubled, Fc is… Fc = mac = mv2/r ?Fc = m(2v)2/r

Fc is 22 or 4x greater!

When a car turns a corner on a level road, which force provides the necessary centripetal acceleration?

A. FrictionB. TensionC. Normal forceD. Air resistanceE. Gravity

Slide 6-8

Answer:When a car turns a corner on a level road, which force provides the necessary centripetal acceleration?

A. FrictionB. TensionC. Normal forceD. Air resistanceE. Gravity

Slide 6-9

A car approaches a level, circular curve with a radius of 45.0 m. If the concrete is dry (µs=0.900), what is the maximum speed at which the car can “negotiate” the curve at a constant speed?

Did you get…???∑F = ma ∑F = Ff = Fc

µsFN = Fc

µsFg = Fc

µsmg = Fc

µsmg = mv2/r (aha… mass cancels) µsg = v2/r v2= µsgr v=√(µsgr) v=√[(0.9)(9.80m/s2)(45m)] v = 19.9 m/s

Example Problem A level curve on a mountain road has a radius of 150 m. What is the maximum speed at which this curve can be safely negotiated on a rainy day when the coefficient of friction between the tires on a car and the road is 0.40?

Did you get…???∑F = ma ∑F = Ff = Fc

µsFN = Fc

µsFg = Fc

µsmg = Fc

µsmg = mv2/r (again… m cancels) µsg = v2/r v2= µsgr v=√(µsgr) v=√[(0.4)(9.80m/s2)(150m)] v = 24.2 m/s

A world-class thrower can get the hammer up to a speed of 29 m/s. If an athlete swings the mass in a horizontal circle centered on the handle he uses to hold the chain, what is the tension in the chain?

In the hammer throw, an athlete spins a heavy mass in a circle at the end of a chain, then lets go of the chain. For male athletes, the “hammer” is a mass of 7.3 kg at the end of a 1.2 m chain.

Example Problem

Hammer Throw… and did you get?m = 7.3 kg, r = 1.2 m, v = 29 m/sFT = ?FT = m∙v2/r = (7.3 kg) ∙ (29 m/s)2 / (1.2 m)

FT = 5116 N {or about 5100 N}

A car of mass 1500 kg goes over a hill at a speed of 20 m/s. The shape of the hill is approximately circular, with a radius of 60 m, as in the figure to the right. When the car is at the highest point of the hill,a.What is the force of gravity

on the car?b.What is the magnitude and

direction of the normal force of the road on the car at this point?

Example Problem: Driving Over A Rise

Apparent weight = the magnitude of the force that supports you (FN).

At top: FN + Fg = Fnet = Fc= mv2/r so, FN = mv2/r - Fg(feel lighter)

At bottom: FN - Fg = Fnet = Fc = mv2/rso, FN = Fg + mv2/r(feel heavier)

Example Problem: Loop-the-LoopA roller coaster car goes through a vertical loop at a constant speed. For positions A to E, rank order the:

• centripetal acceleration• normal force• apparent weight

Answer

The Globe of Death is a spherical cage in which motorcyclists ride in circular paths at high speeds. One outfit claims that riders achieve a speed of 60mph in a 16-ft diameter sphere. 1000 m = 0.621 miles, 1 inch = 2.54 cm, 100 cm = 1m• What would be the period for this motion? • What would be the apparent weight of a

60.0 kg rider at the bottom of the sphere?

Answer