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A uniform solid disk and a uniform hoop are placed S side by side at the top of an incline

of height h.

(a) If they are released from rest and roll without slipping, which object reaches the bottom first?

(b) Verify your answer by calculating their speeds when they reach the bottom in terms of h.

Respuesta :

Manetho

Answer:

Explained

Explanation:

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

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

[tex]\sqrt((9.8\times 1)/0.75)[/tex] = 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 4×Vd= sqrt(4/3hg) Vh= 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)

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