A mallet consists of a uniform cylindrical head of mass 2.18 kg and a diameter D

Question

A mallet consists of a uniform cylindrical head of mass 2.18 kg and a diameter D = 0.074 m mounted on a uniform cylindrical handle of mass 0.520 kg and length L = 0.290 m, as shown in figure below. If this mallet is tossed, spinning, into the air, how far above the bottom of the handle is the point that will follow a parabolic trajectory?
A mallet consists of a uniform cylindrical head of mass 2.18 kg and a diameter D = 0.074 m mounted on a uniform cylindrical handle of mass 0.520 kg and length L = 0.290 m, as shown in figure below. If this mallet is tossed, spinning, into the air, how far above the bottom of the handle is the point that will follow a parabolic trajectory?
A mallet consists of a uniform cylindrical head of mass 2.18 kg and a diameter D = 0.074 m mounted on a uniform cylindrical handle of mass 0.520 kg and length L = 0.290 m, as shown in figure below. If this mallet is tossed, spinning, into the air, how far above the bottom of the handle is the point that will follow a parabolic trajectory?

Explanation / Answer

it is the center of mass that will follow a parabolic trajectory

the coordinate of the center of mass is:

ycm=(m1 y1+m2 y2)/(m1+m2)

where m1, m2 are the masses of the handle, mallet, and y1, y2 are the coordinates of their individual centers of mass

since we are told the mallet and handle are uniform, we can treat them as point masses with the mass concentrated at the center of mass of each individual component

so, the y coordinate of the handle's center of mass is 0.145m; and for the mallet it is 0.037 + 0.29 = 0.327m (since the mallet is mounted at the end of the handle)

the y coordinate of the center of mass for the whole system is:

ycm=(0.520kg x 0.145m + 2.18kg x 0.327m)/(0.520kg+2.18kg)
ycm=0.291 m

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