-Is this all the information it gives you? If it is, then the answer would be the same as the first. The disc of the same radius will reach the bottom at 4m/s.......
So, initial energy = final energy gives us
mv^2 = mgh
v^2 = gh
16 = gh
h = 16/g
Now, let's look at the disk:
Initial energy is still only gravitational potential, or
Initial Energy = U_g = mgh = mg(16/g) = 16m
Final energy is still rotational kinetic and kinetic energy, but our moment of intertia has changed. We also don't know what v is.
Final Energy = (1/2)mv^2 + (1/2)Iw^2
= (1/2)mv^2 + (1/2)(1/2)m(r^2)w^2 = (1/2)mv^2 + (1/4)m(r^2)(w^2)
Subbing in w = v/r, we have
Final Energy = (1/2)mv^2 + (1/4)m(v^2) = (3/4)m(v^2)
Set this equal to initial energy
16m = (3/4)m(v^2)
16 = (3/4)(v^2)
v = sqrt(16*4/3)
Write down the equations of motion of a ring which rolls purely down a slope.
you will get a = gsinθ/2 where θ is the angle of inclination of the slope.
since v^2 - u^2 = 2as
here u=0
v^2=sgsinθ where is length of the inclined plane
v = sqrt(sgsinθ)
4 = sqrt(sgsinθ)
for disc a =2/3gsinθ
v^2=4/3sgsinθ
v = 2sqrt(sgsinθ/3)
since sqrt(sgsinθ) = 4
v = 2*4/sqrt3
v = 8/sqrt3
Yes, jaynar is right. Say if a soccer ball drops from a building in 10 seconds, and you took another soccer ball and dropped it the same way, it would still be 10 seconds. The same applies for your "ring" problem, gravity is the same for all objects.
A spinning ring has more angular momentum than an equally massed spinning disk because its mass is concentrated on its perimeter.
My guess is that the disk will roll faster.
Edit
I beg your pardon "Math" but I think angular momentum is a form of rotational energy.
Is this all the information it gives you? If it is, then the answer would be the same as the first.
The disc of the same radius will reach the bottom at 4m/s.
