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. A cylinder of mass M and radius R is released from rest at the top of an inclined plane. The cylinder rolls without slipping down the incline. The rotational inertia of the cylinder is MR2/2.
Derive an expression for the angular momentum of the cylinder about its center of mass when it has rolled a vertical distance h. Express your answer in terms of M, R, h, and physical constants, as appropriate.

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okpalawalter8

For a cylinder of mass M and radius R is released from rest at the top of an inclined plane, the expression is mathematically given as

L=MR * √{gh/3}

What is the expression for the angular momentum of the cylinder about its center of mass when it has rolled a vertical distance h.?

Generally, the equation for the angular velocity  is mathematically given as

w=V/R

Therefore

[tex]w=\frac{\sqrt{ugh/3}}{R}[/tex]

In conclusion, at instance of IW, the angular velocity

[tex]L=MR^2/2 * 2\sqrt{gh/3}/R[/tex]

L=MR * √{gh/3}

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