## Essential University Physics: Volume 1 (3rd Edition)

We know that the sides are perfectly symmetrical, so the center of mass would normally be in the center of the silo if there were no base. We now must consider the base. The base is at the bottom, so we find the percentage of the mass it takes up and multiply this value by 15, which would normally be the center of mass: $15 \times ( \frac{6000}{44000})=2.04$ Thus, the height of the center of mass is: $15 - 2.04 \approx 13$ Thus, the center of mass is: a) (0,0,13) b) We solve this by adding the mass to the overall silo. We know that the center of mass of the first 44,000 kg is at 13 meters high. We know that the center of mass of the $\pi(2^2)(30)(800)(2/3)=201,061 \ kg$ remaining is 10. Thus, we find the overall center of mass: $h_{cm} = \frac{(201,061\times 10) + (44,000\times 13)}{44,000+201,061}\approx 11$ Thus, the center of mass is: b) (0,0,11)