Answer:
Explanation:
Given
three objects namely Solid sphere, Solid cylinder and hollow cylinder are released from same elevation
Suppose all have same mass and radius
acceleration of object and while rolling
[tex]a=\frac{g\sin \theta }{1+\frac{I}{mr^2}}[/tex]
Moment of inertia of Solid sphere [tex]I_1=\frac{2}{5}Mr^2[/tex]
Moment of inertia of Solid cylinder [tex]I_2=Mr^2[/tex]
Moment of inertia of Hollow cylinder [tex]I_3=\frac{1}{2}Mr^2[/tex]
therefore [tex]a_1=\frac{g\sin \theta }{1+\frac{\frac{2}{5}Mr^2}{Mr^2}}[/tex]
[tex]a_1=\frac{5}{7}g\sin \theta [/tex]
[tex]a_2=\frac{1}{2}g\sin \theta [/tex]
[tex]a_3=\frac{2}{3}g\sin \theta [/tex]
Distance traveled is same so
[tex]s=ut+\frac{1}{2}at^2[/tex]
[tex]u=0[/tex]
[tex]t=\sqrt{\frac{2s}{a}}[/tex]
time is inversely Proportional to square root of acceleration
time Period of solid sphere [tex]t_1=k\cdot 1.183[/tex]
[tex]t_2=k\cdot 1.414[/tex]
[tex]t_3=k\cdot 1.22[/tex]
where k holds for other constant terms
so time taken by Solid sphere is least and last one to reach at bottom is solid cylinder
