Given the situation as shown in the figure. The sphere is rolling down an inclin

Question

Given the situation as shown in the figure. The sphere is rolling down an inclined plane! Mass of the solid sphere is M, radius is R and vertical height is H. V_CM is the linear velocity of the center of mass, I_CM is the moment of inertia and omega is the angular velocity. a. Write energy conservation for the system. b. If I_CM = (2/5) MR^2 and omega = V/R, what will be the velocity, V_CM when ball reaches the bottom? c. If the vertical height, H = 2.8 m and gravitational acceleration, g=9.8m/s^2, calculate the velocity?

Explanation / Answer

(A) Applying energy conservation,

PEi + KEi = PEf + KEf

M g h + 0 = 0 + M v_cm^2 /2 + Icm w^2 /2

M g h = M v_cm^2 /2 + Icm w^2 /2

(b) M g h = M V^2 /2 + (2 M R^2 / 5) (V / R)^2 / 2

M g h = M V^2 / 2 + M V^2 / 5 = 7 M V^2 / 10

V = sqrt[ 10 g h / 7]

(d) V = sqrt(10 x 9.8 x 2.8 / 7) = 6.26 m/s

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

---

---

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