bkhofmann said:Homework Statement
A club-axe consists of a symmetrical 11.0 kg stone attached to the end of a uniform 2.9 kg stick, as shown in the figure. The length of the handle is L1 = 93.0 cm and the length of the stone is L2 = 10.0 cm. How far is the center of mass from the handle end of the club-axe?
Homework Equations
3. The Attempt at a Solution [/b
I'm not sure how to do this, I can up with 58.9cm on my own but most likely not the correct way, any help appreciated!
bkhofmann said:I believe that's what I have done. Do you come up with 58.9 cm?
bkhofmann said:Ok the stone weighs 1.1kg/cm and the stick is 1.1 kg/35.279cm. Is this what you are talking about?
bkhofmann said:I got it (46.5x2.9+98x11)/(2.9+11), the stone is just added on.
bkhofmann said:the center of mass is at 87.26cm from the end of the stick and lon-capa accepted it.
The center of mass is the point in an object where all the mass is evenly distributed around it, making it the balance point of the object.
The center of mass is calculated by taking the sum of the products of each individual mass and its corresponding distance from a chosen reference point, divided by the total mass of the object.
Knowing the location of the center of mass is important because it helps determine the stability and balance of an object. It also helps in predicting the motion of an object and understanding how external forces will affect it.
The location of the center of mass affects the handling of a club-axe by determining the point where the axe will pivot or rotate when swung. If the center of mass is closer to the handle end, it will be easier to control and swing the axe. If it is closer to the blade end, the axe may feel heavier and more difficult to control.
Yes, the location of the center of mass can be changed by altering the distribution of mass in the object. For a club-axe, this could be done by adding weight to one end or removing weight from another, thereby shifting the center of mass closer to one end or the other.