- #1

Alexandra Fabiello

- 42

- 1

- Homework Statement
- A small ice cube is released from rest at the top edge of a hemispherical bowl. When it reaches the bottom of the bowl, its speed is 2.5 m/s. Use conservation of energy to find the radius of the bowl in centimetres. Neglect friction.

- Relevant Equations
- 1/2 mv1^2 + mgy1 = 1/2 mv2^2 + mgy2

v1 = 0 m/s

v2 = 2.5 m/s

y1 - y2 = distance a quarter of the way around the bowl (since we're neglecting friction)

mass can be factored out, so it isn't needed, and some simplifying and the like gets this formula:

v

so 2.5

6.25 = 19.6y

y = 0.318877551 m * 4 = 1.275510204 m = circumference of bowl = 2πr

1.275510204/2π = 0.203003753 m * 100 = 20 cm

But apparently this is wrong, so I don't know what got messed up.

v2 = 2.5 m/s

y1 - y2 = distance a quarter of the way around the bowl (since we're neglecting friction)

mass can be factored out, so it isn't needed, and some simplifying and the like gets this formula:

v

_{2}^{2}= v_{12}+ 2g(y_{1}- y_{2})so 2.5

^{2}= 0 + 2(9.8)(y)6.25 = 19.6y

y = 0.318877551 m * 4 = 1.275510204 m = circumference of bowl = 2πr

1.275510204/2π = 0.203003753 m * 100 = 20 cm

But apparently this is wrong, so I don't know what got messed up.