Question 5.7: The outside diameter of a pulley is 0.8 m, and the cross sec...
The outside diameter of a pulley is 0.8 m, and the cross section of its rim is as shown. Knowing that the pulley is made of steel and that the density of steel is ρ = 7.85 × 10³ kg/m³, determine the mass and weight of the rim.
STRATEGY: You can determine the volume of the rim by applying the second Pappus-Guldinus theorem, which states that the volume equals the product of the given cross-sectional area and the distance traveled by its centroid in one complete revolution. However, you can find the volume more easily by observing that the cross section can be formed from rectangle I with a positive area and from rectangle II with a negative area (Fig. 1).
MODELING: Use a table to keep track of the data, as you did in Sec. 5.1.
| Area, mm² | \pmb{\bar{\text{y}},\text{mm}} | Distance Traveled by C, mm |
Volume, mm³ | |
| I | +5000 | 375 | 2π(375)=2356 | (5000)(2356)=11.78 \times 10^6 |
| II | -1800 | 365 | 2π(365)=2293 | (-1800)(2293)=-4.13 \times 10^6 |
| \text{Volume of rim} = 7.65 \times 10^6 | ||||
ANALYSIS: Since 1 \ \text{mm}=10^{-3}\text{ m}, you have 1\text{ mm}^{-3}=(10^{-3} \text{ m})^3=10^{-9}\text{ m }^3. Thus you obtain V=7.65 \times 10^6 \text{ mm}^3=(7.65 \times 10^6)(10^{-9}\text{ m}^3)=7.65 \times 10^{-3}\text{ m}^3.
m= \rho V=(7.85 \times 10 ^3 \text{ kg/m}^3)(7.65 \times 10^{-3} \text{ m}^3) \quad \quad \quad m=60.0 \text{ kg} \\ W=mg=(60.0\text{ kg})(9.81 \text{ m/s}^2)=589 \text{ kg.m/s}^2 \quad \quad \quad W=589 \text{ N}
REFLECT and THINK: When a cross section can be broken down into multiple common shapes, you can apply Theorem II of Pappus–Guldinus in a manner that involves finding the products of the centroid (\bar{\text{y}}) and area (A), or the first moments of area (\bar{\text{y}}A), for each shape. Thus, it was not necessary to find the centroid or the area of the overall cross section.
