A bike tire with a mass of 65 g spins at a rate of 8 rads per second. What is the moment of inertia for the tire if the radius is 0.24 m? Assume the tire has all its mass concentrated in the rim and that the spokes are mass-less. Round your answer to four decimal places. I used the equation I=MR^2 but it was considered wrong. Please explain. Thank you.


taking the tire as a ring, I = MR² = 0.065•0.24² = 0.003744 kg•m² rounded to 0.0037 kg•m² I is moment of inertia in kg•m² I = cMR² M is mass (kg), R is radius (meters) c = 1 for a ring or hollow cylinder c = 2/5 solid sphere around a diameter c = 7/5 solid sphere around a tangent c = ⅔ hollow sphere around a diameter c = ½ solid cylinder or disk around its center c = 1/12 rod around its center, R = length c = ⅓ for a rod around its end, R = length c = 1 for a point mass M at a distance R from the axis of rotation c = 1/3 for a door, where r is the width for a wheel with r inner radius and R outer radius I = ½M(R²–r²)