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physics

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A 10.0-kg cylinder rolls without slipping on a rough surface. At the instant its center of amss has a speed of 10m/s, determine (a) the translational kinetic of its center of mass (b) the rotational kinetic energy about its center of mass and (c) its total kinetic energy.

can u check if this is the correct answer?

a) KE=.5Mv^2
KE=.5(10)(10)^2
KE=500J

B) KE=.25Mv^2
KE=.25(10)(10)^2
KE=250J

C) 250J+500J=750J

please tell me if that is correct ornot XD thanks

^please check that work i have shown above thanks

  • physics - ,

    this is correct

  • physics - ,

    wrong wrong wrong ..... you need to start reading before you try plugging numbers in like an accountant with a nearing deadline, son ........

  • physics - ,

    According to my physics book with the answers in the back, this is correct.

  • physics - ,

    why are you using .25 for b??

  • physics - ,

    cyndi used 1/4 because of this:
    KErotational=1/2IW^2
    =1/2(1/2mr^2)(V/r)^2
    =1/2(1/2mr^2)(V^2/r^2)
    =1/4mV^2

    as you can see the radii cancel and the 1/2's multiply together to make 1/4

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