- #1
Darkalyan
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Homework Statement
Consider a frictionless sphere of radius r=5.0m that is fixed to the ground and a small block of mass m=1.0 kg that is nudged slightly from the very top. This takes place on the surface of the Earth. Find the location on the ground where the block lands by following the steps below: I have drawn the item using Paint, and I'm attaching an image of it:
a. Find the numerical value of the angle at which the block leaves the sphere.
b. What is the velocity of the block at the instant it leaves the sphere? (Note the components)
c. Find how far from the sphere's point of contact with the ground (0 on the diagram) the block hits the ground.
Homework Equations
F=m*a
g=9.81 m/s^2
KE+PE=constant
KE=.5*m*v^2
W=F*d
W=change in KE
v^2 (final) = v^2 (initial) * 2a*delta X
The Attempt at a Solution
Well, what I thought was that, since there is only a tiny nudge in the horizontal direction, the velocity in that direction is basically ignorable. Thus, the block will accelerate straight downward along the path of the sphere until it hits an angle of 90 degrees, at which point it will continue to fall straight downward.Thus, here is my answer solutions:
A) angle=90 degrees
B) I plug into the equation relating v^2 and 2a*deltaX. In this case, v^2 initial=0, a=9.81, and delta X=5 meters. When those are plugged in, I get the final velocity to be about 9.905 m/s.
C) Because the mass fell at the angle of 90 degrees, it feel straight down, meaning the distance from the point of contact is simply the radius of the sphere, which is 5m.