# physics

posted by .

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