A circular platform of radius Rp = 5.57 m and mass Mp = 315 kg rotates on fricti

Question

A circular platform of radius Rp = 5.57 m and mass Mp = 315 kg rotates on frictionless air bearings about its vertical axis at 7.89 rpm. An 78.9-kg man standing at the very center of the platform starts walking (at t = 0) radially outward at a speed of 0.543 m/s with respect to the platform. Approximating the man by a vertical cylinder of radius Rm = 0.237 m, determine an equation (specific expression) for the angular velocity of the platform as a function of time. What is the angular velocity when the man reaches the edge of the platform?

Explanation / Answer

using angular momentum conservation,

initially = (315 x5.572 /2) x (7.89 x 2/60) = 4037.35

after time t , he wil be at distance fo 0.543t fron center.

then angular moemntum = [(315 x5.572 /2 ) + (78.9 x0.2372 /2   + 78.9 x 0.543t2 )] x

                                 = 4911.9t2

so equating them ,

4037.35 = 4911.9t2

(t)   = 0.822 / t2    ..........Ans

when man reaches to end , then t = 5.57 / 0.543

= 0.8222 / (5.57 / 0.543)2   = 0.00781 rad/s     or 0.0745 rpm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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