The drawing shows a version of the loop-the-loop trick for a small car. If the c

Question

The drawing shows a version of the loop-the-loop trick for a small car. If the car is given an initial speed of 3.54 m/s, what is the largest value that the radius r can have if the car is to remain in contact with the circular track at all times?

Explanation / Answer

its speed can be determined by conservation of energy: ?KE = ?PE m(vi)²/2 - mv²/2 = mgh (vi)² – v² = 2gh (vi)² – v² = 2gr(1–cos?) v² = (vi)²–2gr(1–cos?) At that point, centripetal component of the car's acceleration is: a_c = v²/r a_c = [(vi)²–2gr(1–cos?)] / r a_c = (vi)²/r – 2g(1–cos?) This acceleration is produced by the combination of two forces acting on the car: 1) The normal force F_n (the track pushng inward); 2) The radial component of the gravitational force. Its magnitude is: mgcos?. Sometimes (depending on ?) this is positive (pushing toward loop's center); sometimes negative (pushing away from center); sometimes zero. So, the net force pushing cenripetally inward, as a function of ?, is: F_net_c = F_n – mgcos? (the minus sign reflects the fact that the gravitational component points _outward_ when ? = 0). Now, by F=ma, we have: F_net_c = m(a_c) F_n – mgcos? = m[(vi)²/r – 2g(1–cos?)] or: F_n = m[(vi)²/r – 2g(1–cos?) + gcos?] F_n = m[(vi)²/r – 2g + 3gcos?] Now, we said previously that we want F_n to be greater than or equal to zero for all ?. Therefore the following has to be true for all ?: m[(vi)²/r – 2g + 3gcos?] >= 0 (vi)²/r – 2g + 3gcos? >= 0 (vi)² >= rg(2 – 3cos?) The right side is a maximum when cos? = –1 (namely when the car is at the top of the loop). In that case, we need: (vi)² >= rg(2 – 3(–1)) (vi)² >= 5rg Therefore, the car's inital speed vi needs to be at least sqrt(5rg). 5 years ago Report Abuse 100% 1 Vote Not the right answer? Try Yahoo! Search Search Yahoo! for Search Other Answers (1) zsm28 If the car remains in contact with the track at the highest position, it will do at all times. So we only need to discuss the question there. At the highest position, 2 forces act on the car, namely the gravity mg and the normal force N, both vertically downward. Their net force make the car centripetal acceleration mv^2/r. Frm newton's 2nd law, we have mg + N = mv^2/r N = m(v^2/r - g) N must greater than 0, at least equal 0, the maintain the contact. so v^2/r - g > = 0 r < = v^2/g therefore the largest value of r is v^2/g = 3.9^2/9.8 = 1.6 m

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