A muddy bicycle tire with a radius of 34.0 cm has been tire turns mounted above

Question

A muddy bicycle tire with a radius of 34.0 cm has been tire turns mounted above the ground for cleaning. The vertical distance from the tire axle to the ground is 57.2 cm. r The tire has been given a good spin in the direction shown, with a starting rate of 180 rpm, but thereafter it has been gradually slowing at a rate of .0692 rad/s2. The tire has been spinning for exactly 1 minute when two chunks of mud, located at positions A and B, simultaneously fly off the tire. Each chunk lands on the ground, (touching no other object meanwhile). How much time elapses between their moments of impact with the ground?

Explanation / Answer

Given ?o=6p rad/s, a=-6.92 rad/s², t=60s, s=0.572m, g=9.8m/s² The equation of motion that applies here is: ? = ?o+at = (6p)+(-6.92)(60) = -396 rad/s Which is obviously wrong. The 180rpm seems reasonable, so I'm guessing the "gradual slowing" of 6.92 rad/s² is a typo. If we try 0.692 rad/s, it gives us ?=-23 rad/s, which is better but still wrong. If we try 0.0692 rad/s², it would take 4.5 minutes to stop the wheel which is reasonably gradual. So then we have: ? = 14.7 rad/s. The mud chunk A from the wheel side spinning downward would hit ground first: v = r? = 5m/s The equation of motion that applies here is: s = vt+½gt² 0 = ½gt²+vt-s = 4.9t²+5.0t-0.572 Which can be solved for time t using quadratic equation. ta = 0.10s The mud chunk B from the wheel side spinning upward would hit ground last. Using the same equation of motion and quadratic equation gives us: tb = 2v/g = 1.02s t = tb-ta = 0.92s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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