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deeprony7

  • 3 years ago

A sphere of mass 50gm and radius 10cm rolls without slipping with a velocity of 5cm/s. Its total kinetic energy in ergs is?

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  1. stan
    • 3 years ago
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    For rolling w/o slipping, and choosing teh cm of the cm of teh sphere as the rotation point, \[K _{total}=K _{transl}+K _{rot}=1/2mv^{2}+1/2I \omega ^{2} = 1/2mv^{2}+1/2(2/5mr ^{2})(v/r)^{2}=7/10 mv^{2}\] You can also choose the point of contact as the rotation point and use teh parallel axis theorem to get the same answer

  2. deeprony7
    • 3 years ago
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    how will i do that?

  3. stan
    • 3 years ago
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    Choose the point of contact as the rotation point... thus the point of rotation has been moved a distance r from the center of mass of the sphere. Parallel axis theorem says \[I _{\parallel} = I _{cm} + mr ^{2}\] where, again, r is the distance the point of rotation has been displaced from the center of mass (cm) of the sphere. Now, all energy is rotational: \[K _{total} = K _{rot} = 1/2 I _{\parallel} \omega ^{2} \] \[= 1/2 (2/5 mr ^{2} + mr ^{2})*(v/r)^{2}\] \[=1/2(7/5 mr ^{2})*(v ^{2}/r ^{2}) = 7/10 mv ^{2}\]

  4. deeprony7
    • 3 years ago
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    interesting..=)

  5. dinainjune
    • 3 years ago
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    8.75x10^(-5) Joule

  6. ac7qz
    • 3 years ago
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    You have a combination of linear and rotational KE. These guys are telling you true. P.S. Who uses ergs any more? I haven't heard the term since 1950s sci-fi.

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