A solid sphere is released from the top of a ramp that is at a height A solid sp

Question

A solid sphere is released from the top of a ramp that is at a height

A solid sphere is released from the top of a ramp that is at a height h1 = 1.70 m. It goes down the ramp, the bottom of which is at a height of h2 = 1.25 m above the floor. The edge of the ramp is a short horizontal section from which the ball leaves to land on the floor. The diameter of the ball is 0.18 m. Through what horizontal distance d does the ball travel before landing? How many revolutions does the ball make during its fall?

Explanation / Answer

Here

Total Energy remains COnstant

Therefore

Loss in Potenteial Energy = Gain in Linear Kinetic Energy + Gain in Rotational Kinetic ENERGY

Mg*(h1 - h2) = 0.5*M*v^2 + 0.5*I*w^2

As w = v/r and I = (2/5)mr^2

Mg(h1 - h2) = 0.5*M*v^2 + 0.5*(2/5)Mv^2

Mg(h1 - h2) = (7/10)Mv^2

v = sqrt((10/7)*g*(h1 - h2))

= sqrt((10/7)*9.8*(1.70-1.25))

= 2.51 m/sec

As we know that

h = ut + 0.5*g*t^2

1.25 = 0.5*9.8*t^2

t = 0.505 sec

Therfore

d = vt

= 2.51*0.505

d = 1.2677 m

theta = wt

= (v/r)*t

= (2.51/0.09)*0.505

= 14.08389 rad/sec

No. of revolutions = 14.08389/(2pi)

= 2.24 revolutions

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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