Physics, 07.07.2020 14:01 JustinLaucirica

A uniform solid disk and a uniform ring are place side by side at the top of a rough incline of height h. a) If they are released from rest and roll without slipping, determine the velocity vring of the ring when it reaches the bottom.
b) Verify your answer by calculating their speeds when they reach the bottom in terms of h.

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Answer from: nanagardiner08

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 $=\sqrt((gh)/0.7)$

Answer from: Quest

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Answer from: Quest

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