Calculate the rotational inertia of a meter stick, with mass 0.82 kg, about an a

Question

Calculate the rotational inertia of a meter stick, with mass 0.82 kg, about an axis perpendicular to the stick and located at the 32 cm mark. (Treat the stick as a thin rod.) Next, A 34.0 kg wheel, essentially a thin hoop with radius 1.05 m, is rotating at 280 rev/min. It must be brought to a stop in 15.0 s.(a) How much work must be done to stop it? (b) What is the required average power? Also Calculate the rotational inertia of a wheel that has a kinetic energy of 24,400 J when rotating at 685 rev/min.

Explanation / Answer

1) (0.82*0.32*0.16*0.16/3) + (0.82*0.68*0.34*0.34/3) = 0.002239+0.0214= 0.0236 2) T * 1.05 =- 34*1.05*1.05* 30.787/15 T = troque = - 76.936 ,,,3) work done to stop = - 0.5* 34*1.05*1.05* 30.787*30.787 = -17764.879 J 4) 0.5* I *75.319*75.319 = 24400 I = 8.6022

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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