Asked by cyndi

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

Answers

Answered by Angie
this is correct
Answered by fo' sho
wrong wrong wrong ..... you need to start reading before you try plugging numbers in like an accountant with a nearing deadline, son ........
Answered by Kaleb
According to my physics book with the answers in the back, this is correct.
Answered by vm
why are you using .25 for b??
Answered by Erick
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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