Suppose that a solid ball, a solid disk, and a hoop all have the same mass and the same radius. Each object is set rolling without slipping up an incline with the same initial linear (translational) speed. Which object goes farthest up the incline?

Respuesta :

Answer:

Hoop will reach the maximum height

Explanation:

let the mass and radius of solid ball, solid disk and hoop be m and r  (all have same radius and mass)

They all  are rolled with similar initial speed v

by the law of conservation of energy we can write

[tex]K_{trans}+K_{rot}= P[/tex]

for solid ball

[tex][tex]\frac{1}{2}mv^2+\frac{1}{2}I_{ball}\omega^2= mgh_{ball}[/tex]

putting [tex]I_{ball}=\frac{2}{5}mr^2 and \omega=\frac{v}{r}[/tex] in the above equation and solving we get

[tex]h_{ball}= 0.071v^2[/tex]

now for solid disk

[tex][tex]\frac{1}{2}mv^2+\frac{1}{2}I_{disk}\omega^2= mgh_{disk}[/tex]

putting [tex]I_{ball}=\frac{1}{2}mr^2 and \omega=\frac{v}{r}[/tex] in the above equation and solving we get

[tex]h_{disk}= 0.076v^2[/tex]

for hoop

[tex][tex]\frac{1}{2}mv^2+\frac{1}{2}I_{hoop}\omega^2= mgh_{hoop}[/tex]

putting [tex]I_{hoop}=mr^2 and \omega=\frac{v}{r}[/tex] in the above equation and solving we get

[tex]h_{hook}= 0.10v^2[/tex]

clearly from the above calculation we can say that the Hoop will reach the maximum height

The hoop will be able to reach the maximum height

What will be the condition of the given objects?

let the mass and radius of solid ball, solid disk, and hoop be m and r  (all have the same radius and mass)

They all  are rolled with similar initial speed v

by the law of conservation of energy, we can write

[tex]K_{translation }+K_{rotaion}=P[/tex]

for solid ball

[tex]\dfrac{1}{2} mv^2+\dfrac{1}{2}I_{ball}w^2=mgh_{ball}[/tex]

putting   [tex]I_{ball}=\dfrac{2}{5} mr2 \ and \ w=\dfrac{v}{r}[/tex]  in the above equation and solving we get

h[tex]h_{ball}=0.071v^2[/tex]

now for solid disk

[tex]\dfrac{1}{2} mv^2+\dfrac{1}{2}I_{disk}w^2=mgh_{disk}[/tex]

putting [tex]I_{disk}=\dfrac{1}{2} mr^2\ and \ w=\dfrac{v}{r}[/tex] in the above equation and solving we get

[tex]h_{disk}=0.076v^2[/tex]

for hoop

[tex]\dfrac{1}{2} mv^2+\dfrac{1}{2}I_{hoop}w^2=mgh_{hoop}[/tex]

putting [tex]I_{hoop}=mr^2\ and \ w=\dfrac{v}{r}[/tex]  in the above equation and solving we get

[tex]h_{hoop}=0.10v^2[/tex]

By comparing the height from all three it is clear that the hoop has the maximum height.

Thus the hoop will be able to reach the maximum height

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