A 2.5 g ice flake is released from the edge of a hemispherical bowl whose radius r is 40.0 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 a second flake with twice the mass was substituted, what would its speed be?
Should I start by using KE=(1/2)mv^2? I am not really sure how I should approach this.
3 answers
I have so far converted 2.5g to 0.0025kg and 40.0cm to 0.4m.
please disregard this, I've figured it out
An energy method is a good way to get the speed at the bottom.
At the bottom, (1/2) M V^2 = M g R (the P.E. loss).
Note that mass M cancels out
V= sqrt (2 g R)
a and b have the same answer.
The TIME it takes to reach the bottom is a lot harder to calculate, and requires solving a messy differential equation.
At the bottom, (1/2) M V^2 = M g R (the P.E. loss).
Note that mass M cancels out
V= sqrt (2 g R)
a and b have the same answer.
The TIME it takes to reach the bottom is a lot harder to calculate, and requires solving a messy differential equation.