In this diagram, the cylindrical can is on a horizontal surface and AC is a diameter of the base. Also, B is on the top edge of the can and BC is vertical. If the radius of the base is 5cm and the volume is 1178cm3, can you work out the angle between the rod and the base of the can?
Answer: This tin can was not everyone’s cup of tea with only a select handful of readers correctly working out the answer at 56,31°. Extra well done, so, to: Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Michele Girardi, Scame Mastaf Spa, Suisio, Italy; Jose Padron, material development specialist, Waterville TG Inc. Waterville, Québec, Canada; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands. Honourable mentions also to Simon Robinson, editor, Urethanes Technology International, London, UK and Peter McKenzie Vass, technical director, Kerry Abrasives. Listowel, Co. Kerry, Ireland, who were on the right track.
Tan(alpha) = 15/10 (10 being the diameter on the base and 15 the height of the can) = 56,31°.
Volume = (1/2AC)^2*PI*BC
1178=5^2*PI*BC leading to BC = 15
angle BAC= inv tg(1.5)= 56.3
With a volume of 1178 cm³ and radius 5 cm, we can get height of the can.
V= A*H; H=V/A and A= PI*r²
H= 1178 cm³/(PI*5²) = 14.999 cm = BC
Diam. Base = 2r = 2*5 = 10 cm = AC
Thus; Tangent of angle BAC = BC/AC = 14.999/10 = 1.4999
Angle between rod and base = Arc Tan (1.4999) = 56.31
Diameter base A-C is 10 cm, area of base s (Pi)*r2 = 3.1415926*25 = 78.54 cm2 If volume is 1178 cm2, then height B-C is 1178/78.54 = 14.99876 cm
Angle between A-B and A-C is deduced from tan (a) = B-C/ A-C = 15/10 = 1.499876
From tables angle (a) is approx. 56,4 degrees.
New teaser on Monday