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
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
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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