Question

moment of inertia and center of mass velocity question.

Answer 1

A thin light string is wrapped around the outer rim of a uniform hollow cylinder of mass 4.15 kg having an inner and outer radii of 20 cm and 35 cm. The object starts from rest. How far must the cylinder fall before it's center is moving at 7 m/s? If you just dropped this cylinder without any string, how fast would it's center be moving when it had fallen the previous distance found when attached to the string? Why do you get two different answers?

Answer 2

first i used this\[mgh=\frac{1}{2}mv_{com}^2+\frac{1}{2}I\omega^2\]and got this\[mgh=|\frac{1}{2}mv_{com}^2+\frac{1}{2}\left[\frac{1}{2}m(R_1^2+R_2^2)\left( \frac{v_{com}^2}{R_2^2}\right)\right]\]and then got h=264.11

Answer 3

did i do this right?

Answer 4

What are the units for h?

Answer 5

it's in meters since i just converted the radius inter meters.

Answer 6

Does the answer seem reasonable to you?

Answer 7

nope...

Answer 8

agreed...so what assumption did you make when you used the conservation of energy?

Answer 9

well honestly... i don't even know what's gong on. I'm just using my notes i had from class. we did this exact problem but it was a sphere so i guess i didn't make any assumptions other than assuming that this would be almost the same as the sphere.

Answer 10

ok...hang on a sec.

Answer 11

ok so if you are confident the physics is correct....could you have made a math mistake?

Answer 12

i'm not sure. i just erased everything and i'm going to start from scratch again so i'll see if get something different. don't wait up on me though. lol

Answer 13

I will wait...I did a quick calculation in my head and got a very different answer than you....

Answer 14

like 262 meters different?

Answer 15

h = (.5v^2 + 1/4 (R1^2 + R2^2)V^2/r2^2)/g...
I got about 4 meters....

Answer 16

and you are now getting....

Answer 17

sorry. still working

Answer 18

no problem...

Answer 19

ok i'm doing it step by step and i get some huge number again. i don't know what i'm doing wrong. the second time i got 250.39

Answer 20

Try this...h = (.5(49) + 1/4(.20^2 + .35^2)*49/.35^2)/9.8 =

Answer 21

wait. i think i see what i did wrong. cripes. this part. (R1^2+R2^2)(v_com^2/R2^2) i just divided out the R2 from R2 not from R1 so instead i ended up with (R1+1)(v_com^2) instead of (R1^2/R2^2+1)(v_com^2)

Answer 22

I just did the calculation on my abacus and got 4.16 meters

Answer 23

I will be back on tonight so we should discuss this so you know why conservation of energy works here

Answer 24

ok i get the same as you. 4.15816

Answer 25

ok thank you

Answer 26

no problem at all