Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

a small cylinder of radius r is released coaxially from point A inside the fixed large cylindrical bowl of radius r as in fug.if the friction between the small and large cylinder is sufficient enough t o prevent any slipping then find (a) the fractions of total KE vs rotational when the cylinder reaches the bottom .(b) the normal force exerted by small cylinder on the larger one when it is at bottom.

Physics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

|dw:1328285800230:dw|
do you use conservation law of E? between A& bottom point
?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

in point A it has PE & in bottom has KE (rotational & transitional KE) do you get my goal?
\[mgr=(1/2mv ^{2})+(1/2)I \omega ^{2}\]
ys
then?
for sylender \[I=(1/2mr ^{2})\]
wait please...
from 2nd law of newton have \[N-mg=mv ^{2}/R\] right?
in part 1 question ask this: \[(0.5mv ^{2}+0.5\omega ^{2})/(0.5I \omega ^{2})\] so v=Rw i think ithink answer is:2[(R/r)]^2
sorry one of the i think is lampoon
big cylender is frictionless ok?
i answered part b on top

Not the answer you are looking for?

Search for more explanations.

Ask your own question