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A group of students perform the same "Conservation of Mechanical Energy" experiment that you performed in lab by allowing a solid sphere and then a solid cylinder to roll down the ramp. Both objects were released from the same position at the top of the ramp. If the speed vsphere of the solid sphere at the bottom of the ramp was 1.25 m/s, what would be the speed vcylinder at the bottom of the ramp?

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Answer:

[tex]V_ {cylinder} = 1.11m / s[/tex]

Explanation:

For this case we simply find the speeds for both

in the situations mentioned as well,

[tex]V_ {sphere} = \sqrt {\frac {10gh} {7}}[/tex]

[tex]1.15 = \sqrt {\frac {10 (9.8) h} {7}}[/tex]

Solving for h,

[tex]h = 0.0944m[/tex]

[tex]V_ {cylinder} = \sqrt {\frac {4gh} {3}}[/tex]

[tex]V_ {cylinder} = \sqrt {\frac {4 (9.8) (0.0944)} {3}}[/tex]

[tex]V_ {cylinder} = 1.11m / s[/tex]

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