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Chapter 8

Q. 8.6

SYSTEM OF RODS

GOAL Find the center of mass of a continuous system of particles.

PROBLEM A rod system with a uniform linear density of 5.00 \mathrm{~kg} / \mathrm{m} is cast into an irregular, upside-down T-shape as in Figure 8.13 \mathrm{a}, with rod 1 the horizontal rod and rod 2 the vertical rod. Neglect the diameter of the rods. (a) Find the mass of each rod. (b) Determine the center of mass of each rod. (c) Calculate the center of mass of the system in the x y-plane.

STRATEGY For part (a), the mass of each rod is given by m=\lambda L, where \lambda is the mass per unit length and L is the length. The center of mass of each of the rods, part (b), can be found by symmetry: it’s in the middle of the rod. In this case, inspection suffices, but otherwise the midpoint formula could be applied. Finally, the center of mass of the system is the same as if the mass of each rod were concentrated at that rod’s center of mass, in which case Equations 8.3 \mathrm{a}

x_{\mathrm{cg}}=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3} + \cdots }{m_{1}+m_{2}+m_{3} + \cdots } =\frac{\sum m_{i} x_{i}}{\sum m_{i}}    [8.3a]

and 8.3 \mathrm{b}

y_{\mathrm{cg}}=\frac{\sum m_{i} y_{i}}{\sum m_{i}}      [8.3b]

can be applied.

8.13

Step-by-Step

Verified Solution

(a) Find the mass of the each rod.

Use the linear density and substitute, finding the mass of the horizontal rod, m_{1} :

m_{1}=\lambda L_{1}=(5.00 \mathrm{~kg} / \mathrm{m})(3.00 \mathrm{~m})=15.0 \mathrm{~kg}

Do the same for the mass of the vertical rod, m_{2} :

m_{2}=\lambda L_{2}=(5.00 \mathrm{~kg} / \mathrm{m})(2.00 \mathrm{~m})=10.0 \mathrm{~kg}

(b) Find the center of mass of each rod.

Because each rod has uniform density, the center of mass is in the center of each rod:

CM, horizontal rod: \left(x_{1}, y_{1}\right)=(1.50,0) \mathrm{m}

CM, vertical rod: \left(x_{2}, y_{2}\right)=(1.00,1.00) \mathrm{m}

(c) Find the center of mass of the system in the x y-plane.

Apply Equations 8.3 \mathrm{a} and 8.3 \mathrm{b} :

\begin{aligned}&x_{\mathrm{cm}}=\frac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}}=\frac{(15.0 \mathrm{~kg})(1.50 \mathrm{~m})+(10.0 \mathrm{~kg})(1.00 \mathrm{~m})}{15.0 \mathrm{~kg}+10.0 \mathrm{~kg}}=1.30 \mathrm{~m} \\&y_{\mathrm{cm}}=\frac{m_{1} y_{1}+m_{2} y_{2}}{m_{1}+m_{2}}=\frac{(15.0 \mathrm{~kg})(0 \mathrm{~m})+(10.0 \mathrm{~kg})(1.00 \mathrm{~m})}{15.0 \mathrm{~kg}+10.0 \mathrm{~kg}}=0.400 \mathrm{~m}\end{aligned}

REMARKS The center of mass is outside the system at the position calculated. Note that knowledge of the linear density was not really required because it essentially occurs in both the numerator and denominator, and would cancel out. See the exercise.