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Chapter 7: Q62P (page 170)

A mallet consists of a uniform cylindrical head of mass 2.30 kg and a diameter 0.0800 m mounted on a uniform cylindrical handle of mass 0.500 kg and length 0.240 m, as shown in Fig. 7–42. 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?

FIGURE 7-42 Problem 62.

Short Answer

Expert verified

The point that will follow a parabolic trajectory above the bottom of the handle is \(25.1\;{\rm{cm}}\).

Step by step solution

01

Define translational motion

The translation motion is an object motion that occurs if each object element travels an equal distance in the same direction.

A car moving in a straight line is an example of translation motion.

02

Given information

The mass of the head is\({m_{\rm{h}}} = 2.30\;{\rm{kg}}\).

The mass of the handle is \({m_{\rm{H}}} = 0.500\;{\rm{kg}}\).

03

Step 3: Calculate the position of the center of mass of the handle and head

The position of the center of mass of the handle can be calculated as shown below:

\(\begin{array}{c}{x_{\rm{H}}} = \frac{{24.0\;{\rm{cm}}}}{2}\\{x_{\rm{H}}} = 12.0\;{\rm{cm}}\end{array}\)

The position of the center of mass of the head can be calculated as shown below:

\(\begin{array}{c}{x_{\rm{h}}} = \left( {24.0\;{\rm{cm}}} \right) + \frac{{\left( {8.00\;{\rm{cm}}} \right)}}{2}\\{x_{\rm{h}}} = 28.0\;{\rm{cm}}\end{array}\)

04

Step 4: Calculate the center of mass of the handle and head system

The point above the bottom of the handle that follows the trajectory is the center of mass of the handle and head system. The location of this point is

\(\begin{array}{c}{x_{{\rm{CM}}}} = \frac{{{m_{\rm{H}}}{x_{\rm{H}}} + {m_{\rm{h}}}{x_{\rm{h}}}}}{{{m_{\rm{H}}} + {m_{\rm{h}}}}}\\{x_{{\rm{CM}}}} = \frac{{\left( {0.500\;{\rm{kg}}} \right)\left( {12.0\;{\rm{cm}}} \right) + \left( {2.30\;{\rm{kg}}} \right)\left( {28.0\;{\rm{cm}}} \right)}}{{\left( {0.500\;{\rm{kg}}} \right) + \left( {2.30\;{\rm{kg}}} \right)}}\\{x_{{\rm{CM}}}} = 25.1\;{\rm{cm}}{\rm{.}}\end{array}\)

Thus, the point that follows a parabolic trajectory above the bottom of the handle is \(25.1\;{\rm{cm}}\).

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