Question

Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 7.6 m/s. Ignore frictional losses. (a) What is the height of the hill? (b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational speed of the frozen juice can when it reaches the bottom?

Answer 1

(A) h = 4.9 m v

(B) = 8 m/s

Step-by-step explanation:

translational speed = 7.6 m/s

(A) from the conservation of energy equation

mgh = (mv^{2} / r) + (Iw^{2} / r)

where the moment of inertial (I) = (2mr^{3}) / 3

mgh = \frac{mv^{2}}{r} + \frac{\frac{2mr^{3}}{3} x w^{2} }{r}

mgh = m (\frac{v^{2}}{r} + \frac{\frac{2r^{3}}{3} x w^{2} }{r} )

gh = \frac{5V^{2}}{6}

g = \frac{5V^{2}}{6g}

h = \frac{(5) x (7.6)^{2}}{6 x 9.8}

h = 4.9 m

(B) just as from in (A) above, using the conservation of energy equation with (I) being close to mr^2 / 2

mgh = (mv^{2} / r) + (1/2)(mr^{2}/2)(v/r)^{2}

v =
`\sqrt{(4gh)/(3) }`

v =
`\sqrt{(4 x 9.8 x 4.9)/(3) }`

v = 8 m/s