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
5 answers
this is correct
wrong wrong wrong ..... you need to start reading before you try plugging numbers in like an accountant with a nearing deadline, son ........
According to my physics book with the answers in the back, this is correct.
why are you using .25 for b??
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
2 answers