A bicycle rim has a diameter of 0.65 m and a moment of inertia, measured about its center, of 0.25 kg⋅m2. part a what is the mass of the rim? express your answer to two significant figures and include the appropriate units. m =

2.4 kg The moment of inertia of a body is the sum of the mass of every particle in the body multiplied by the square of the distance of that particle from the axis of rotation. With that definition in mind, the radius of the rim will be 0.65 m / 2 = 0.325 m. And the moment of inertia will be expressed as I = MR^2 So substitute the known values into the expression and solve for m, giving: I = MR^2 0.25 kg*m^2 = m * (0.325 m) ^2 0.25 kg*m^2 = M * 0.105625 m^2 0.25 kg*m^2 = M * 0.105625 m^2 2.366863905 kg = M Rounding to 2 significant figures gives 2.4 kg