- #1
EndoBendo
- 11
- 0
A block sits at the top of a smooth sphere of radius 1m. Suddenly, under the
force of gravity, It begins to slide down the surface of the sphere until it leaves the
surface. At what speed does it leave the surface?
mgcosθ + N = mv^2/R
But N =0
v^2 = Rgcosθ
V^2 = gcosθ
speed = v = ( 9.8cosθ)^0.5
Now,
from work energy theorem,
0.5mv^2 - 0 = mgR(1-cosθ)
Rgcosθ = 2gR(1-cosθ)
cosθ = 2(1-cosθ)
3cosθ = 2
cosθ = 2/3
Now
SPeed = v = (9.8 x 2/3)^0.5 = 2.556 m/s
I am unsure if there's supposed to be a 1/2 in front of the mv^2/R on the first line of my work, which would change the final value to 6.32 m/s
force of gravity, It begins to slide down the surface of the sphere until it leaves the
surface. At what speed does it leave the surface?
mgcosθ + N = mv^2/R
But N =0
v^2 = Rgcosθ
V^2 = gcosθ
speed = v = ( 9.8cosθ)^0.5
Now,
from work energy theorem,
0.5mv^2 - 0 = mgR(1-cosθ)
Rgcosθ = 2gR(1-cosθ)
cosθ = 2(1-cosθ)
3cosθ = 2
cosθ = 2/3
Now
SPeed = v = (9.8 x 2/3)^0.5 = 2.556 m/s
I am unsure if there's supposed to be a 1/2 in front of the mv^2/R on the first line of my work, which would change the final value to 6.32 m/s
