Approximate the track by a circular loop of radius 6.70 m . Assume that Steve\'s

Question

Approximate the track by a circular loop of radius 6.70 m . Assume that Steve's mass is 81.90 kg , and assume that the mass of the car is 1058.0 kg (not including Steve). Also assume that the central axis of the track points east and west, and that the car is going north at the start and at the end of the stunt. The questions may also ask you to use the descriptive radial and tangential coordinate system, with the car's direction of motion being tangentially counterclockwise (CCW).

Assume that once Steve enters the loop he keeps his feet away from both the brake pedal and the gas pedal. If so, and if air resistance, friction in the wheel bearings, and the rotational inertia of the wheels can be ignored (this is a technical way of saying that we don't yet wish to think about the forces that are needed to make the wheels turn at a faster or slower rate), then it is easy to calculate the speed of the car with height above the ground. Recal the following equation from your study of constant acceleraton a in the y direction: =2avAy (1) For the case of free fall, and if posiive y corresponds to UP, then ay has a value of negative 9.8 mis2 G.e. negative g). Also for free fall, with z as the horizontal direction, the z velocity never changes, so that - 0 (2) Adding equations (1) and (2) and solving for the final speed squared yelds the following equation which is true for the case of free fall: where the Theorem of Pythagoras has been used on the component velocities to produce the speeds. The remarkable fact is that this free-fall equation also works for the loop-the-loop, as long as the assumptions listed above are reasonable. You will earn why this equation works when you study energy methods, but for now simply use it (equation (3)) to calculate the speed for various heights. hen using equation (3), remember that positive y corresponds to UF Forces at the Bottom of the Loop Suppose that, for the actual stunt, Stove decides to risk a speed of 10.00 m/s at the top of loop. Part D For a spood of 10.00 m/s at the top of loop (i.o. for the actual stunt), what is tho car's spood just after it has entered the circular loop but before tho track has risen to any significant hoight abavo the ground? Use equation (3) m/'s Submit Part E What is the magritude of the normal force on Steve by the carsoat at this point in the stunt (e.just after entoring the curved track but bofore the track has risen to any significant height above the groundy? View Available Hint(s) 31?

Explanation / Answer

D)

Vf^2 = Vi^2 + 2gy

10^2 = Vi^2 + 2(-9.8)(2x6.7) ... diference in height between top and bottom is equal to diameter

Vi = 19.04 m/s speed while entering

E)

At this point, the weight and centrifugal force act downwards, combining to form the action and forming equal and opposite reaction by the car seat

R = mg + mv^2 / r

= 81.9 x 9.8 + 81.9 x 19.04^2 / 6.7

= 5234 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.