A solid, uniform ball of mass 17.4 kg, and radius 0.32 m, rolls without slipping

Question

A solid, uniform ball of mass 17.4 kg, and radius 0.32 m, rolls without slipping up a hill of height 27.4 m, as shown in the figure . At the bottom of the hill the ball has a velocity of 29.7 m/s. At the top of the hill, it is moving horizontally, and then it goes over the vertical cliff.
part A. What is the total initial kinetic energy of the ball at the bottom of the hill? Don't forget to use the numbers in the introduction not the picture.
part B. What is the velocity of the ball just as it leaves the top of the hill?
part C. How long is the ball in the air?
part D. How far from the foot of the cliff does the ball land?
part E. How fast is it moving just before it lands?

A solid, uniform ball of mass 17.4 kg, and radius 0.32 m, rolls without slipping up a hill of height 27.4 m, as shown in the figure . At the bottom of the hill the ball has a velocity of 29.7 m/s. At the top of the hill, it is moving horizontally, and then it goes over the vertical cliff. part A. What is the total initial kinetic energy of the ball at the bottom of the hill? Don't forget to use the numbers in the introduction not the picture. part B. What is the velocity of the ball just as it leaves the top of the hill? part C. How long is the ball in the air? part D. How far from the foot of the cliff does the ball land? part E. How fast is it moving just before it lands?

Explanation / Answer

Given:
mass of the ball = 17.4 kg
radius of the ball = 0.32 m
velocity ofthe ball at the bottom of the hill = 29.7 m/s.
_____________________________________________________________________________

(a)
Moment of inertia of sphere = 2/5MR^2
By conservation of mechanical energy
ki+ui = kf+uf-------(1)
here the sphere has kinentic energy so initial potential energy is zero.
hence above equaion can be reduces as
ki = kf+uf
the sphere has both rotational and translation motion then
(1/2I^2)+(1/2Mvi^2) = Mgh+5If^2+(1/2Mvf^2)

so the above eqution reduces as

(1/5vi^2)+(1/2vi^2) = gy+(1/5vf^2)--------(2)

(7/10)vi^2 = gh+(7/10)vf^2

vi = 29.7 m/s and h = 27.4 m

from equation 2

final velocity vf = 22.3 m/s.at the top of the hill.

___________________________________________________________________________________

(b)

initial velocity of the the ball just it leaves the top of the hill is the final velocit y of the ball.

v = 22.3 m/s

___________________________________________________________________________

(c)

height h = s = 27.4 m

initial velocity = 22.3

time of flight = sqrt(2h/g)

                      = sqrt(2* 27.4/9.8)

                      = 2.36sec

________________________________________________________________________

(d)

range R = u sqrt(2h/g)

              = 22.3(2.36)

              = 52.6 m

The fall leaving at the top of the cliff follows the projectile motion so we consider

the range.

_______________________________________________________________________________

(e)

v = sqrt(2gh)

   = sqrt(2*9.8*26.4)

   = 22.74 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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