Question

A large sphere rolls without slipping across a horizontal surface as shown. The sphere has a constant translational speed of 10. m/s, a mass of 7.3 kg (16 lb bowling ball), and a radius of 0.20 m. The moment of inertia of the sphere about its center is I = (2/5)mr². The sphere approaches a 25° incline of height 3.0 m and rolls up the ramp without slipping.

(a) Calculate the total energy, E, of the sphere as it rolls along the horizontal surface.

(b) Does the sphere have enough energy to make it to the top of the ramp? You must show work for credit.

(c) If the sphere does not make it to the top, how far does it travel up the ramp?  If it does make it, where does it land, as measured from the base of the ramp?

(d) Suppose, instead, that the sphere were to roll toward the incline as stated earlier, but this time the ramp is frictionless. Explain what will happen to the sphere under this new condition.

Answer 1

a )

translational speed of 10. m/s = V

m = mass of 7.3 kg

r = radius of 0.20 m

The moment of inertia of the sphere about its center is I = (2/5) m r²

= 25° incline of

h = height 3.0 m

KE = 1/2 m v² + 1/2 I ²

= v / r

and I = 2/5 m r²

KE = 1/2 m v² + 1/2 I ²

= 1/2 m v² + 1/2 ( 2/5 m r²) (v/r)²

KE = ( 7/10 ) m v²

= 0.7 X 7.3 X 10²

KE = 511 J

b )

KE = PE = m g h + 1/2 m v²

511 = 7.3 X 9.8 X 3 + 0.5 X 7.3 X v²

296.38 = 0.5 X 7.3 X v²

v = 9.0111 m/s

direction is = 25° above the horizontal

c )

v_x = 9.0111 X cos25 = 8.166 m/s

v_y = 9.0111 X sin25 =3.808 m/s

h = v_y t + 1/2 at²

- 3 = 3.808 X t - 4.9 X t²

4.9 X t² - 3.808 t - 3 = 0

t = 1.262 sec

distance d = v_x t

= 8.166 X 1.262

d = 10.3 m

d )

it is greater than value

because the ball only sliding but not rolling.