Respuesta :
Answer:
Step-by-step explanation:
Given
Hoop, Uniform Solid Cylinder, Spherical shell and a uniform Solid sphere released from Rest from same height
Suppose they have same mass and radius
time Period is given by
[tex]t=\sqrt{\frac{2h}{a}}[/tex] ,where h=height of release
a=acceleration
[tex]a=\frac{g\sin \theta }{1+\frac{I}{mr^2}}[/tex]
Where I=moment of inertia
a for hoop
[tex]a=\frac{g\sin \theta }{1+\frac{mr^2}{mr^2}}[/tex]
[tex]a=\frac{g\sin \theta }{2}[/tex]
a for Uniform solid cylinder
[tex]a=\frac{g\sin \theta }{1+\frac{mr^2}{2mr^2}}[/tex]
[tex]a=\frac{2g\sin \theta }{3}[/tex]
a for spherical shell
[tex]a=\frac{g\sin \theta }{1+\frac{2mr^2}{3mr^2}}[/tex]
[tex]a=\frac{3g\sin \theta }{5}[/tex]
a for Uniform Solid
[tex]a=\frac{g\sin \theta }{1+\frac{2mr^2}{5mr^2}}[/tex]
[tex]a=\frac{5g\sin \theta }{7}[/tex]
time taken will be inversely proportional to the square root of acceleration
[tex]t_1=k\sqrt{2}=1.414k[/tex]
[tex]t_2=k\sqrt{\frac{3}{2}}=1.224k[/tex]
[tex]t_3=k\sqrt{\frac{5}{3}}=1.2909k[/tex]
[tex]t_4=k\sqrt{\frac{7}{5}}=1.183k[/tex]
thus first one to reach is Solid Sphere
second is Uniform solid cylinder
third is Spherical Shell
Fourth is hoop
