In Fig. 8-32, a 1.9 g ice flake is released from the edge of a hemispherical bow
ID: 2175706 • Letter: I
Question
In Fig. 8-32, a 1.9 g ice flake is released from the edge of a hemispherical bowl whose radius r is 35 cm. The flake-bowl contact is frictionless. (a) What is the speed of the flake when it reaches the bottom of the bowl? (b) If we substituted a second flake with 4 times the mass, what would its speed be? (c) If, instead, we gave the flake an initial downward speed 1.6 m/s along the bowl, what would the answer be?
In Fig. 8-32, a 1.9 g ice flake is released from the edge of a hemispherical bowl whose radius r is 35 cm. The flake-bowl contact is frictionless. (a) What is the speed of the flake when it reaches the bottom of the bowl? (b) If we substituted a second flake with 4 times the mass, what would its speed be? (c) If, instead, we gave the flake an initial downward speed 1.6 m/s along the bowl, what would the answer be?Explanation / Answer
This is done using the conservation of energy.
Part A)
At the rim of the bowl the flake has only PE. At the bottom it has only KE.
PE = KE
mgh = .5mv2 (Note mass cancels, so it does not matter how massive the flake is)
gh = .5v2
v = (2gh) where h is the radius of the of the bowl - that is how far down the flake falls
v = [(2)(9.8)(.35)]
v = 2.62 m/s
Part B)
Since mass cancels, the same formula applies to this portion since the flake still starts from rest. Thus the answer here is the same as in part A.
v = 2.62 m/s
Part C)
For this part, the flake will have some initial KE as well as PE, thus
KEi + PE = KEf
.5mvi2 + mgh = .5mvf2 (again, mass cancels)
.5(1.6)2 + (9.8)(.35) = (.5)(vf2)
vf = 3.07 m/s