What Is the Acceleration Ratio of a Disk to a Hoop Rolling Down an Incline?

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A uniform solid disk and a uniform hoop rolling down an incline from rest exhibit different speeds at the bottom, with the disk achieving a velocity of sqrt(4/3hg) and the hoop sqrt(hg). The discussion centers on determining the ratio of their accelerations, a_disk to a_hoop. The calculations reveal that the acceleration ratio is 4/3, derived from the relationship between their final velocities and the distance traveled. The importance of rolling without slipping is emphasized, as it affects the time taken to reach the bottom. Ultimately, the correct ratio of their accelerations is confirmed to be 4/3.
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hoop and disk rolling(help please)

ok I've been looking at this forever and can't get it...dont know what to do..i would appreciate it if someone could help

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A uniform solid disk and a uniform hoop are
placed side by side at the top of an incline of
height h.
If they are released from rest and roll with-
out slipping, determine their speeds when
they reach the bottom. (here i found the speeds to be
velocity disk = sqrt(4/3hg)
velocity hoop = sqrt(hg)

now here is my problem

What is the ratio of their accelerations as they
roll down the incline, a_disk / a_hoop?

1. 4/3
2. 2
3. sqrt(3)
4. sqrt(3/2)
5. 1/2 xxxx
6. sqrt(4/3)
7. 1/3 xxxx
8. 3
9. 3/2
10. sqrt(2)

..ps - the ones i put an xxxx out the right on i figured couldn't be it since the velocity of the disk is greater the acceleration of the disk(top value) must be greater
 
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Mac13 said:
If they are released from rest and roll with-
out slipping, determine their speeds when
they reach the bottom. (here i found the speeds to be
velocity disk = sqrt(4/3hg)
velocity hoop = sqrt(hg)

now here is my problem

What is the ratio of their accelerations as they
roll down the incline, a_disk / a_hoop?

1. 4/3
2. 2
3. sqrt(3)
4. sqrt(3/2)
5. 1/2 xxxx
6. sqrt(4/3)
7. 1/3 xxxx
8. 3
9. 3/2
10. sqrt(2)

Write the acceleration for each
disk: sqrt(4/3hg)/t
hoop: sqrt(hg)/t

Now divide them and see what you get.
disk/hoop = sqrt(4/3) which is answer 6

You remember how to divide radicals right?
 
Last edited:
ShawnD's answer isn't quite correct. Because the two objects are rolling down with a different acceleration, there is a difference in time in order to get to the bottom. However, regardless, we do know the equation

v_f^2 = v_i^2 + 2ad
In case LaTeX doesn't work, v_f^2 = v_i^2 + 2ad,

both initial velocities are zero, so we solve for a in terms of v_f and get

a = \frac{v_f^2}{2d}
a = v_f^2/(2d)

Since d is the same for both, we can compare these two velocities, so our answer will come from

a1/a2 = (v_f1/v_f2)^2 = (Sqrt[4/3hg]/Sqrt[hg])^2 = Sqrt[4/3]^2 = 4/3

cookiemonster
 
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