Question

An empty beer can has a mass of 50 g, a length of 12 cm, and a radius of 3.3 cm. Assume that the shell of the can is a perfect cylinder of uniform density and thickness.
(a) What is the mass of the lid/bottom?
(b) What is the mass of the shell?
(c) Find the moment of inertia of the can about the cylinder's axis of symmetry.

Answer 1

(a) The mass has three components, M = M_lid + M_shell + M_bottom = 50 g, where the lid and the bottom are identical. Mass is proportional to the surface area for uniform objects. The area of the lid and bottom is that of a circle, A_circle = ?r² . The surface area of a cylindrical shell is A_shell = 2?rL. So the total area of can is A_total = 2A_circle + A_shell = 2?rL + 2?r² = 2?r(r + L). So the mass of the lid or bottom is given by $S07_0008.gif$ . Thus M_lid = 5.392 g.

(b) Similarly, the mass of the shell is given by $S07_0009.gif$ . Thus M_shell = 39.216 g.

(c) The total moment of inertia of the beer can is given by the sum of the individual pieces, I_total = I_lid + I_shell + I_bottom. Since each piece is revolving about its centre of mass, we do not need the parallel axis theorem. Looking up the moments of inertia of a flat solid disk and a thin cylindrical shell, we have I_total = 2׽M_lidr² + M_shellr² = 4.86

Answer 2

Answer 3

An empty beer can has a mass of 50 g, a length of 12 cm, and a radius of 3.3 cm. Assume that the shell of the can is a perfect cylinder of uniform density and thickness.
(a) What is the mass of the lid/bottom?
(b) What is the mass of the shell?
(c) Find the moment of inertia of the can about the cylinder's axis of symmetry.

$S07_0004.gif$

(a) The mass has three components, M = M_lid + M_shell + M_bottom = 50 g, where the lid and the bottom are identical. Mass is proportional to the surface area for uniform objects. The area of the lid and bottom is that of a circle, A_circle = ?r² . The surface area of a cylindrical shell is A_shell = 2?rL. So the total area of can is A_total = 2A_circle + A_shell = 2?rL + 2?r² = 2?r(r + L). So the mass of the lid or bottom is given by $S07_0008.gif$ . Thus M_lid = 5.392 g.

(b) Similarly, the mass of the shell is given by $S07_0009.gif$ . Thus M_shell = 39.216 g.

(c) The total moment of inertia of the beer can is given by the sum of the individual pieces, I_total = I_lid + I_shell + I_bottom. Since each piece is revolving about its centre of mass, we do not need the parallel axis theorem. Looking up the moments of inertia of a flat solid disk and a thin cylindrical shell, we have I_total = 2׽M_lidr² + M_shellr² = 4.86

Answer 4

(a) The mass has three components, M = M_lid + M_shell + M_bottom = 50 g, where the lid and the bottom are identical. Mass is proportional to the surface area for uniform objects. The area of the lid and bottom is that of a circle, A_circle = ?r² . The surface area of a cylindrical shell is A_shell = 2?rL. So the total area of can is A_total = 2A_circle + A_shell = 2?rL + 2?r² = 2?r(r + L). So the mass of the lid or bottom is given by $S07_0008.gif$ . Thus M_lid = 5.392 g.

(b) Similarly, the mass of the shell is given by $S07_0009.gif$ . Thus M_shell = 39.216 g.

(c) The total moment of inertia of the beer can is given by the sum of the individual pieces, I_total = I_lid + I_shell + I_bottom. Since each piece is revolving about its centre of mass, we do not need the parallel axis theorem. Looking up the moments of inertia of a flat solid disk and a thin cylindrical shell, we have I_total = 2׽M_lidr² + M_shellr² = 4.86