Understanding Center of Mass Distribution in Rotational Equilibrium

[omarMihilmy]

Isn't the point of center of mass is where the masses on both sides are equally distributed?

Why is it not the case here?

 

[SteamKing]

What do you mean by 'equally distributed'?

 

[omarMihilmy]

I mean that if i have this pencil

The right side's mass is equal to the left hand's side length!

 

[adjacent]

Isn't the point of center of mass is where the masses on both sides are equally distributed?

The bat is wider at the top and thinner at the bottom.So the C.G will not be in the middle of the bat.

I don't understand what yu mean here.Masses on both sides should equal.
The masses on both sides of C.G is same,so your answer is wrong.

 

[omarMihilmy]

Okay great my answer is c) both sides are equal the books answer is b) that the right side is greater why is that?

 

[adjacent]

I mean that if i have this pencilView attachment 65062
The right side's mass is equal to the left hand's side length!

Length?

Right side's mass is equal to the left side's mass.I think it's a typo.Please correct it

 

[omarMihilmy]

Serway would never have a Typo it clearly is correct for some reason

 

[adjacent]

Serway would never have a Typo it clearly is correct for some reason

All the members of PF would deny that "If i have a pencil,the right side's mass is equal to the left hand's side length!"

http://en.wikipedia.org/wiki/Center_of_mass

 

[omarMihilmy]

There is a misunderstanding !

 

[omarMihilmy]

Okay let's start fresh from the original question

The bat has a center of mass closer the left side and I am asked to see if the masses are equal or if one side is greater than the other?

I have an understanding that the center of mass divides the system into two parts of equal mass. Is this true or not? If not then how do I know from the picture given in my first post that which side has a greater mass?

 

[SteamKing]

I have an understanding that the center of mass divides the system into two parts of equal mass. Is this true or not? If not then how do I know from the picture given in my first post that which side has a greater mass?

The center of mass most emphatically does not divide a body into two parts having equal mass. The location of the center of mass depends on how the mass is distributed within the body itself.

I understand your confusion, however. With bodies like a pencil or a uniform steel bar, yes, the amount of mass on either side of the c.o.m. will be equal, but this result applies only to such bodies where the mass is distributed uniformly and evenly about an axis of symmetry and is not a general rule.

The baseball bat, unfortunately, does not have its mass distributed evenly over its length.

What the center of mass does, however, is divide a body such that the first moment of mass on either side of the c.o.m. is equal.

 

[iRaid]

The weight of the baseball bat is not equal in all spots. Since the bat is wider at one end, it weighs more on that end. That's the reason why the center of mass of an object with some density can be represented as an integral as the volume changes.

 

[haruspex]

The weight of the baseball bat is not equal in all spots. Since the bat is wider at one end, it weighs more on that end. That's the reason why the center of mass of an object with some density can be represented as an integral as the volume changes.

No, that misses the point. SteamKing's answer is the correct one.
Consider for example two baseball bats joined in a line, the thin end of one being joined to the fat end of the other. By your reasoning, the CoM could be at the join, but it isn't.

 

[Yanick]

SteamKing is correct about the the torque being equal (but opposite) on both sides of a COM, and not necessarily the masses. Imagine joining the two ends with a massless bar of some sort and placing a triangle underneath that massless bar such that the two halves are in translational and rotational equilibrium. Now imagine the COM of each piece and remember that the clockwise torque has to equal the counter clockwise torque to maintain rotational equilibrium. The COM of the fat end will be closer to the pivot whereas that of the thin end is further from the pivot. For the torques to be equal, the mass on the right (fat end) has to be greater to make up for a shorter length to the pivot as compared to the thin end.