Asked by Anonymous
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.
(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.
Answers
Answered by
Anonymous
I have so far converted 2.5g to 0.0025kg and 40.0cm to 0.4m.
Answered by
Anonymous
please disregard this, I've figured it out
Answered by
drwls
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.