# Center of Mass...Really?!

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#### SaintJude

##### Full Member
10+ Year Member
This question is from the Kaplan Test review, Physics Subject Test 1.

A cylindrically shaped beam of wood 12 ft long & 2 feet in diameter lies length wise. If its weight of 200 pounds is uniformly distributed, what is the minimum amount of work needed to lift the beam such that it stands on one end?

Answer ends up having everything to do with center of mass. If anyone like I can post the answer steps, but how was I supposed to know center of mass was involved?? Any tips?

#### milski

##### 1K member
10+ Year Member
Since the body is at rest both before and after you're done lifting it, the work done is going to be the difference between the potential energies. The easy way to determine potential energy (from gravity) of a body is to use its center of mass and how high it is above a reference 0.

#### Danlee07

##### Full Member
10+ Year Member
And you can do the above calculation since the questions says "uniformly distributed."

#### 1stmeds

##### started from the bottom
10+ Year Member
Would you mind posting the answer steps so I can check my answer and reasoning.

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#### ljc

##### Full Member
10+ Year Member
Localize this object's mass at it's center. Now we have a point of mass weighing 200lbs.

On it's side, the center of mass is at a height of 1 ft (center of diameter, which is 2ft). We'll treat this as Y=0, because we're finding a difference. Potential energy is PE=mgh=200*10*0 (except convert lbs to kgs and ft to meters, which I'm too lazy to do)

On it's end, the center of mass is at 6 ft (center of height, which is 12ft). This is now Y=5 (6ft is 5ft higher than 1ft). PE=200*10*5 (again, convert units)

To find the amount of work done, just find the change in potential energy. This is the definition of work. Difference in PE's:
(200*10*5)-(200*10*0) = 10000J-0J = 10000J (if units were kg and meters) =10 KJ

#### milski

##### 1K member
10+ Year Member
Adding the conversions, so that people can compare to the numerical answer:

200 lb * 5 ft * 10 m/s^2 = 90.6 kg * 1.524 m * 10 m/s^2 = 1370 J

#### 1stmeds

##### started from the bottom
10+ Year Member
awesome, thank you!

#### 1stmeds

##### started from the bottom
10+ Year Member
...but...although that is how I reasoned it, ljc says the definition of work is difference in PE but isn't work = to the difference in KE? or am I misunderstanding the concept?

#### milski

##### 1K member
10+ Year Member
...but...although that is how I reasoned it, ljc says the definition of work is difference in PE but isn't work = to the difference in KE? or am I misunderstanding the concept?

difference in KE = total work done by all forces
difference in PE in the problem = work done by the lifting force on the object
The same amount of work but with opposite sign was done on the object by the gravity force. When you add them, the total work done ends up zero, which is consistent with no change in KE.

#### 1stmeds

##### started from the bottom
10+ Year Member
Ah, got it. thanks for clarifying