Object 1 has a mass of 99.2 kg. Object 2 has a mass of 42.3 kg. If the 2 object pust against each other, what will be the speed
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To solve this problem we will apply the principle of conservation of energy, for which the initial potential and kinetic energy must be equal to the final one. The final kinetic energy will be transformed into rotational and translational energy, so the mathematical expression that approximates this deduction is
KE_i+PE_i = KE_{trans}+KE_{rot} +PE_f
, since initially cylinder was at rest
since at the ground potential energy is zero
The mathematical values are,
Here,
m = mass
g= Gravity
h = Height
V = Velocity
moment of Inertia in terms of its mass and radius
Angular velocity in terms of tangential velocity and its radius
Replacing the values we have that
mgh = \frac{1}{2} mv^2 +\frac{1}{2} (\frac{mr^2}{2})(\frac{v}{r})^2
gh = \frac{v^2}{2}+\frac{v^2}{4}
v = \sqrt{\frac{4gh}{3}}
From trigonometry the vertical height of inclined plane is the length of this plane for , then
Replacing,
Therefore the cylinder's speedat the bottom of the ramp is 3.32m/s