# Hoop and disk rolling(help )

1. Mar 25, 2004

### Mac13

hoop and disk rolling(help plz)

ok i've been looking at this forever and cant 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 couldnt be it since the velocity of the disk is greater the acceleration of the disk(top value) must be greater

2. Mar 26, 2004

### ShawnD

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: Mar 26, 2004
3. Mar 26, 2004

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