Twenty problems in three tiers. Answers at the end.
⭐ 1 (§6.2) — 1000 kg car in a curve r=50 m at v=15 m/s. Centripetal force?
⭐ 2 (§6.2) — Curve r=100 m, \mu_s=0.5. Max speed?
⭐ 3 (§6.2) — Vertical loop r=8 m. Min speed at top?
⭐ 4 (§6.3) — Radial a=3, tangential a=4 m/s². Total?
⭐⭐ 5 (§6.2) — Rope L=1.5 m, ball at end at v=3 m/s on horizontal disk. Angle of rope from vertical?
⭐⭐ 6 (§6.2) — Banked turn 10°, radius 200 m. Design speed?
⭐⭐ 7 (§6.5) — Skydiver 70 kg, A=0.7 m², C_d=0.7. Terminal velocity?
⭐⭐ 8 (§6.4) — Car accelerating 3 m/s². Pendulum hanging from ceiling. Angle from vertical?
⭐⭐ 9 (§6.2) — Carousel with T=6 s. Room radius 4 m. \mu_s needed for passenger not to slip?
⭐⭐ 10 (§6.2) — Satellite at 2R_E from center. Orbital speed? (GM = gR_E^2)
⭐⭐ 11 (§6.5) — Raindrop 2 mm diameter. Which regime? Estimate terminal velocity.
⭐⭐ 12 (§6.2) — Train in vertical circular tunnel. At bottom, normal force in units of weight if v = 2\sqrt{gr}?
⭐⭐⭐ 13 (§6.4) — Bus in curve r=100 m at v=15 m/s. Ball on the floor. From the passenger's view, net fictitious force?
⭐⭐⭐ 14 (§6.3) — Pendulum speed v at bottom. Rope tension? (length L, mass m)
⭐⭐⭐ 15 (§6.5) — Fall with F_\text{drag} = bv. Time to reach 90% of terminal velocity?
⭐⭐⭐ 16 (§6.2) — Banked turn in rain, low friction. Maximum speed?
⭐⭐⭐ 17 (§6.4) — Stone released from a 1000 m peak in northern hemisphere. Coriolis deflection estimate?
⭐⭐⭐ 18 (§6.2 + §6.5) — Skydiver at terminal velocity. If she opens the chute, initial acceleration?
⭐⭐⭐ 19 (§6.2) — Car on banked turn θ=15°, r=300 m, \mu_s = 0.3. Max and min speed without slipping?
⭐⭐⭐ 20 (§6.6) — Ratio of electromagnetic to gravitational force between two protons?
Answers
1) F = mv²/r = 4500 N
2) v \approx 22.1\ m/s
3) v_\text{min} = 8.85\ m/s
4) 5\ m/s²
5) \theta \approx 30°
6) v \approx 18.6\ m/s
7) v_t \approx 53\ m/s
8) \theta \approx 17°
9) \mu_s \geq 0.45
10) v = 5590\ m/s
11) Inertial regime. Estimate: ~6-7 m/s
12) N = 5mg. Five times weight.
13) a_c = 2.25\ m/s²; fictitious centrifugal force 2.25m outward. Floor friction can hold it.
14) T = m(g + v²/L)
15) t = 2.3\tau (with \tau = m/b).
16) With very low \mu, only design speed works.
17) Eastward deflection: \sim 0.7\ m east.
18) Initial deceleration \sim -g (upward).
19) v_\text{min} \approx 19 m/s, v_\text{max} \approx 42 m/s.
20) F_e/F_g \approx 1.24 \times 10^{36}. EM is 10^{36} times stronger.
Preview of §6.9
Q&A.
📚 See also: Halliday Vol 1, Ch 6 — Problems.
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