A uniform spherical shell of mass M = 4.2 kg and radius R = 7.6 cm can rotate ab

Question

A uniform spherical shell of mass M = 4.2 kg and radius R = 7.6 cm can rotate about a vertical axis on frictionless bearings (see figure below). A massless cord passes around the equator of the shell, over a pulley of rotational inertia I = 3.0 times 10^-3 kg middot m^2 and radius r = 5.0 cm, and is attached to a small object of mass m = 0.60 kg. There is no friction on the pulley's axle; the cord does not slip on the pulley. What is the speed of the object after it has fallen 78 cm after being released from rest? Use energy considerations. m/s

Explanation / Answer

initial energy Ei = Ui = m*g*h

after falling h = 78 cm = 0.78 m

let v be the speed of mass

angular speed of sphere wsphere = v/R

angular speed of pulley wpulley = v/r

final energy Ef = KEsphere + KEpulley + KEmass

KEsphere = (1/2)*Isphere*wsphere^2

KEsphere = (1/2)*(2/3)*M*R^2*v^2/R^2 = (1/3)*M*v^2

KEpulley = (1/2)*Ipulley*wpulley^2 = (1/2)*I*v^2/r^2

KEmass = (1/2)*m*v^2

from conservation of energy

Ef = Ei

(1/3)*M*v^2 + (1/2)*I*v^2/r^2 + (1/2)*m*v^2 = m*g*h

((1/3)*4.2*v^2) + ((1/2)*3*10^-3*(v^2/0.05^2)) + ((1/2)*0.6*v^2) = 0.6*9.8*0.78

speed v = 1.412m/s <<<------answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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