A uniform solid 5.25-kg cylinder is released from rest and rolls without slipping down an inclined plane inclined at 18° to the horizontal. Its speed while rolling down is 3.0 m/s.
Given:
The mass of the solid cylinder, [tex]m=5.25 kg[/tex]
The inclination of the plane, θ=[tex]18^{0}[/tex]
The speed of a solid cylinder rolling down an inclined surface is calculated using the following formula:
The kinetic energy of the cylinder
[tex]K=3/5 mgh[/tex]
[tex]1/2mv^{2} =3/5 mgh[/tex]
[tex]v=\sqrt{6gh/5}[/tex]
Substituting for the values h as [tex]h=mgcos[/tex]θ
we get the speed as
[tex]v=3.0 m/s[/tex]
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