A thin disc of mass M and radius R has mass per unit area σ r = k r 2 where r is the distance from its centre. Its moment inertia about an axis going through its centre of mass and perpendicular to its plane is:

Question

A thin disc of mass M and radius R has mass per unit area &#963; r = k r 2 where r is the distance from its centre. Its moment inertia about an axis going through its centre of mass and perpendicular to its plane is:

Aicharya · Accepted Answer

d m = &#963; &#8901; d A = &#963; &#8201; 2 &#960; r &#8201; d r = 2 &#960; k &#8201; r 3 &#8201; d r M = &#8747; d m = 2 &#960; k &#8747; 0 R r 3 d r = 2 &#960; k R 4 4 = &#960; k R 4 2 d I = d m &#8201; r 2 = 2 &#960; k r 5 &#8201; d r I = 2 &#960; k &#8747; 0 R r 5 d r = 2 &#960; k R 6 6 = 2 &#960; k R 6 3 &#215; 2 since, M = &#960;kR 4 2 so, I = 2 M R 2 3

A thin disc of mass M and radius R has mass per unit area σ r = k r 2 where r is the distance from its…

Solution

A thin disc of mass M and radius R has mass per unit area &#963; r = k r 2 where r is the distance from its…

Solution

A thin disc of mass M and radius R has mass per unit area σ r = k r 2 where r is the distance from its…