Part B A singlescoop ice cream cone is a composite body made

Part B

A single-scoop ice cream cone is a composite body made from a single scoop of ice cream placed into a cone. (Figure 2) Assume that the scoop of ice cream is a sphere with radius r=1.44in that is placed into a 4.00 in tall cone. The interior height of the cone is 3.60 in . The cone has an exterior radius of 1.25 in and an interior radius of 1.10 in . The scoop of ice cream sits on the cone\'s interior radius and extends into the cone some distance. Find the z¯ centroid for the cone (the scoop of ice cream and the cone).

Part C

The specific weight of the cone and scoop of ice cream are cone=10.0 lb/ft3 and ice cream=45.0 lb/ft3, respectively. What is z¯, the location of the center of gravity of the cone (i.e., the cone and scoop of ice cream)?

Solution

As,

M*z = M1*z1 + M2*z2

where, zi = Centroid of i^th part

and, Mi = Mass of i^th part

As, M1 = Mass of Sphere (Ice Cream) = Volume*Density = (4/3)pi*r³*Density

As, r = radius of cone = 1.44 in

and, Density = 45 lb/ft³ = 0.02604 lb/in³

So, M1 = 0.2452 lb

and, z1 = Centroid of Ice-Cream or Sphere = 4 + r = 5.44 inch

>> Now, Considering Cone

M2 = Mass = Volume*Density = (1/3)*pi*R²*h*Density

As, R = Radius of Cone = 1.25 in

h = height of cone = 4 in

Density of Cone = 10 lb/ft³ = 5.787*10^-3 lb/in³

=> M2 = 0.03788 lb

and, z2 = h/3 = 4/3 = 1.333 inch

So, Mz = M1*z1 + M2*z2

As, M = M1 + M2 = 0.28308 lb

=> z = (0.2452*5.44 + 0.03788*1.33)/0.28308

=> z= Z Coordinate of Centroid = 4.727 inch