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Two uniform solid balls, one of radius R and mass M, the other of radius 2R and mass 8M, roll down a high incline. They start together from rest at the top of the incline. Which one will reach the bottom of the incline first?

Respuesta :

Answer:Both arrive at the same time.

Explanation:

Let [tex]I[/tex] be the moment of inertia.

Let [tex]k[/tex] be the radius of gyration

Let [tex]m[/tex] be the mass of the ball.

Then,it takes time proportional to [tex]\frac{I}{mr^{2}}[/tex].

Moment of inertia of a solid sphere is given by [tex]\frac{2}{5}mr^{2}[/tex]

Ball 1:

[tex]m=M[/tex]

[tex]r=R[/tex]

[tex]I=\frac{2}{5}MR^{2}[/tex]

So,time is proportional to [tex]\frac{\frac{2}{5}MR^{2} }{MR^{2}}=\frac{2}{5}[/tex]

Ball 2:

[tex]m=8M[/tex]

[tex]r=2R[/tex]

[tex]I=\frac{2}{5}(8M)(2R)^{2}[/tex]=[tex]\frac{64}{5}MR^{2}[/tex].

So,time is proportional to [tex]\frac{\frac{64}{5}MR^{2} }{(8M)(2R)^{2}}=\frac{}{}=\frac{2}{5}[/tex]

So,both arrive at the same time.

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