91Ó°ÊÓ

For a hollow sphere: \(I = \frac{2}{3}mr^2\) For a ring: \(I = mr^2\) For a disc: \(I = \frac{1}{2}mr^2\) For a solid sphere: \(I = \frac{2}{5}mr^2\) Since m = 1 and r = 1, we can simplify these expressions: Hollow sphere: \(I = \frac{2}{3}\) Ring: \(I = 1\) Disk: \(I = \frac{1}{2}\) Solid sphere: \(I = \frac{2}{5}\)

We know the angle of inclination is 37 degrees, with \(\tan 37^\circ = \frac{3}{4}\). To find the acceleration for each object, we'll use the formula: \(a = \frac{g\sin{\theta}(1 - \mu k)}{1 + k}\) Here, \(k = \frac{I}{mr^2}\) and g = 10 m/s². Using the calculated values of moment of inertia, we can find the acceleration for each object: Hollow sphere: \(a = \frac{10 \times \frac{3}{5}(1 - \mu \frac{2}{3})}{1 + \frac{2}{3}}\) Ring: \(a = \frac{10 \times \frac{3}{5}(1 - \mu)}{1 + 1}\) Disk: \(a = \frac{10 \times \frac{3}{5}(1 - \mu \frac{1}{2})}{1 + \frac{1}{2}}\) Solid sphere: \(a = \frac{10 \times \frac{3}{5}(1 - \mu \frac{2}{5})}{1 + \frac{2}{5}}\)

Now we have the expressions for the acceleration of each object. To compare the accelerations, we need to analyze these expressions with respect to the coefficient of friction, µ. If µ = 0 (no friction), all objects will have the same acceleration, which is: \(a = \frac{10 \times \frac{3}{5}}{1 + k}\) Solid sphere has the smallest value of k, followed by the disk, hollow sphere, and ring. As µ increases, the frictional force becomes more significant, and the accelerations of the objects will be affected differently. In general, the solid sphere's acceleration will be less affected by friction than the other objects, leading to a higher acceleration than the others. In conclusion, when comparing the accelerations of the four objects on the inclined plane, we can say that the solid sphere will have the highest acceleration, followed by the disk, hollow sphere, and ring. This order may vary depending on the value of the coefficient of friction, µ.