A disk and a sphere of same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first?

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Toppr Admin · Accepted Answer

A-g-xA0-sin-x3B8-1-K2R2For disc- K2R2-12-0-5For sphere- K2R2-25-0-4Since- a-sphere- -gt- a-disc- -x21D2- Sphere will reach first

A disk and a sphere of same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first?
Disk
Sphere
Both reach at the same time
Depends on their masses

$a = \frac{g s i n θ}{1 + \frac{K^{2}}{R^{2}}}$

For disc: $\frac{K^{2}}{R^{2}} = \frac{1}{2} = 0.5$

For sphere: $\frac{K^{2}}{R^{2}} = \frac{2}{5} = 0.4$

Since, a(sphere) > a(disc) $\Rightarrow$ Sphere will reach first

A disk and a sphere of same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first?DiskSphereBoth reach at the same timeDepends on their masses

a=g sinθ1+K2R2For disc: K2R2=12=0.5For sphere: K2R2=25=0.4Since, a(sphere) > a(disc) ⇒ Sphere will reach first

A disk and a sphere of same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first?
Disk
Sphere
Both reach at the same time
Depends on their masses

$a = \frac{g s i n θ}{1 + \frac{K^{2}}{R^{2}}}$

For disc: $\frac{K^{2}}{R^{2}} = \frac{1}{2} = 0.5$

For sphere: $\frac{K^{2}}{R^{2}} = \frac{2}{5} = 0.4$

Since, a(sphere) > a(disc) $\Rightarrow$ Sphere will reach first