Question

A uniform spherical shell of mass M = 4.5 kg and radius R = 8.5 cm can rotate about a vertical axis on frictionless bearings. A massless cord passes around the equator of the shell, over a pulley of rotational inertia I = 3.0 × 10−3 kg · m2 and radius r = 5.0 cm, and is attached to a small object of mass m = 0.60 kg. There is no friction on the pulley’s axle; the cord does not slip on the pulley. What is the speed of the object when it has fallen 82 cm after being released from rest? Use energy considerations.

Answer 1

I think the change in energy would be mgh = 0.6*9.8*0.82

I am not sure yet how to setup rest of the problem...

Answer 2

I understand, here we can write the subsequent equation:
\[\Large mgh = \frac{1}{2}{I_{sp}}\omega _{sp}^2 + \frac{1}{2}{I_{pu}}\omega _{pu}^2 + \frac{1}{2}m{v^2}\]
where \(sp\) stand for sphere, and \(pu\) stands for pulley

Answer 3

stands*

Answer 4

Ahh so if I interpret correctly, that potential energy is being converted to kinetic energy of the object and also the rotational kinetic energy of sphere and pulley

Answer 5

I can fetch \(I_{sp}\) and \(I_{pu}\) from the table
need to find \(\omega_{sp}\) and \(\omega_{pu}\) hmm..

Answer 6

Correct!
work done by gravity is responsible of the motion of our system of pulley+sphere+block furthermore, we can write the subsequent condition
\[\Large {\omega _{sp}}R = {\omega _{pu}}r = v\]
where \(\large v\) is the magnitude of the velocity of the block

Answer 7

since our cable is not extensible

Answer 8

Wow, nice! that linear speed is same through out the cord!
so 3 equations and 3 unknowns, feed to wolfram ? :D

Answer 9

ok! :) usually I work without calculators :)

Answer 10

I think that the literal solution is very good anyway

Answer 11

I am just trying to understand how to setup these problems and getting excited whenever my answer matches with the number on back of textbook... so let me feed it to mother wolfram and see if it gives the correct answer xD

Answer 12

ok! I'm waiting for your result :)

Answer 13

Looks we have it !

http://www.wolframalpha.com/input/?i=solve+0.6*9.8*0.82+%3D+1%2F2*%282*4.5*0.085%5E2%2F3%29*s%5E2+%2B+1%2F2*%283*10%5E%28-3%29%29*p%5E2+%2B+1%2F2*0.6*v%5E2%2C+v+%3D+0.085*s%2C+v+%3D+0.05*p%2C+v%3E0

Answer 14

matches with the textbook XD
https://s3.amazonaws.com/upload.screenshot.co/3f7bf5a0f7

Answer 15

great! :)

Answer 16

thank you @Michele_Laino :)

Answer 17

:)