Question

A uniform solid disk and a uniform ring are place side by side at the top of a rough incline of height h.

Answer 1

Explanation:

velocity of disc $=\sqrt((gh)/0.75)$

lets call (h) 1 m to make it simple.

= 3.614 m/s

$\sqrt((4/3) x 1 x 9.8) = 3.614$ m/s pointing towards this:

$4×V_d=\sqrt(4/3hg)$

$V_h=\sqrt(hg)$

velocity of hoop=$\sqrt(gh)$

lets call (h) 1m to make it simple again.

$\sqrt(9.8 x 1) = 3.13$ m/s

$\sqrt(gh) = sqrt(hg)so [tex]4×V_d= \sqrt(4/3hg)V_h=\sqrt(hg)$

The disc is the fastest.

While i'm on this subject i'll show you this:

Solid ball $=0.7v^2= gh$

solid disc $= 0.75v^2 = gh$

hoop $=v^2=gh$

The above is simplified from linear KE + rotational KE, the radius or mass makes no difference to the above formula.

The solid ball will be the faster of the 3, like above i'll show you.

solid ball: velocity $=\sqrt((gh)/0.7)$

let (h) be 1m again to compare.

$\sqrt((9.8 x 1)/0.7) = 3.741$ m/s

solid disk speed $=\sqrt((gh)/0.75)$

uniform hoop speed $=\sqrt(gh)$

solid sphere speed $=\sqrt((gh)/0.7)$