Asked by Mary
A 2.0 kg disc rolls without slipping on a horizontal surface, so that its center of mass has a constant linear speed of 6.0 m/s. What is the total kinetic energy of the disc?
i know that k= 0.5mv2 + 0.5 I w2
K=1/2 mv2 (1+(I/mr2)
but i am still getting he wrong answer because i convert I to miri2 so therefore it cancels with the denominator and k = 1/2mv2 but that is wrong. i am stuck
i know that k= 0.5mv2 + 0.5 I w2
K=1/2 mv2 (1+(I/mr2)
but i am still getting he wrong answer because i convert I to miri2 so therefore it cancels with the denominator and k = 1/2mv2 but that is wrong. i am stuck
Answers
Answered by
drwls
I for a solid disc is (1/2) m r^2
Try it with that.
If you use I = m r^2 (which would be correct for a hoop), you would get
total KE = mv^2, not(1/2) m v^2
Try it with that.
If you use I = m r^2 (which would be correct for a hoop), you would get
total KE = mv^2, not(1/2) m v^2
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