# Chemical Forums

• March 19, 2019, 09:53:00 PM
• Welcome, Guest

•

Pages: [1]   Go Down

### AuthorTopic: Conservation of Energy Problem (hopefully the last one!)  (Read 5290 times) !function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0];if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src="https://platform.twitter.com/widgets.js";fjs.parentNode.insertBefore(js,fjs);}}(document,"script","twitter-wjs"); (function() {var po = document.createElement("script"); po.type = "text/javascript"; po.async = true;po.src = "https://apis.google.com/js/plusone.js";var s = document.getElementsByTagName("script")[0]; s.parentNode.insertBefore(po, s);})();

0 Members and 1 Guest are viewing this topic.

#### nj_bartel

• Sr. Member
• Mole Snacks: +76/-42
• Offline
• Posts: 1488
##### Conservation of Energy Problem (hopefully the last one!)
« on: October 22, 2008, 05:56:37 PM »

An Atwood's machine consists of two masses, m1 = 5 kg and m2 = 8 kg, which are connected by a light string that passes over a pulley.  The pulley has a mass of 10 kg, a radius of 0.2 m, and a moment of inertia of 0.12 kg m2.  If the 5 kg mass is initially on the floor and the 8 kg mass is at a height of 2.0 m, what speed will the heavier mass have just before hitting the floor?

________________________________________________________________________________________________

I first drew my free body diagrams and set PEi = KEf.

All of the PEi of the system comes from the height of the heavier block, so

PEi = (2 m)(8 kg)(9.81 m/s2)

I then equated this to the KEf of the system -

(1/2)(m1 + m2 kg)(v m/s)2 + (1/2)(I kg m2)(omega rad/s)2

which with the numbers plugged in comes to

(1/2)(13 kg)(v m/s)2 + (1/2)(0.12 kg m2)(omega rad/s)2

This is where I ran into difficulty.  I'm assuming you can't add as is, due to the units being different.  I can easily convert omega into vt by

However, I'm not sure how to convert I into kg's (or do I just use the mass of the entire pulley?)  I tried converting I into kg's by (0.12) / (.2)2, but this came out to 3 kg, and the mass of the pulley is 10 kg.

Thanks,
Nick
Logged

#### Borek

• Mr. pH
• Deity Member
• Mole Snacks: +1628/-395
• Offline
• Gender:
• Posts: 24864
• I am known to be occasionally wrong.
##### Re: Conservation of Energy Problem (hopefully the last one!)
« Reply #1 on: October 22, 2008, 09:17:08 PM »

Radian doesn't have physical units. By definition it is m/m - so it is 1. Compare Holiday and Resnick, in my (Polish) edition that's the first footnote in Rotational motion section.

Haven't you forgot something about the other mass?
Logged
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info, PZWT_s1

#### nj_bartel

• Sr. Member
• Mole Snacks: +76/-42
• Offline
• Posts: 1488
##### Re: Conservation of Energy Problem (hopefully the last one!)
« Reply #2 on: October 23, 2008, 03:11:15 AM »

Oh!  Can't believe I was treating radians as units.  So I only have to convert angular velocity into tangential velocity, correct?  To allow me to solve for v.

Which other mass are you talking about?  If it's the PE of the pulley, that's going to be constant and I can disregard it, can I not?  If you're talking about the mass of the lighter weight, it can be added to the larger weight mass to get 13, can't it?  I'm not sure what you're getting at =/