Consider a circular hoop, a disk, and a sphere all having the same mass and radius. These objects all start at rest at the top of a ramp and are released to roll down the ramp at the same instant. If they all roll straight down the ramp without slipping or colliding, which one reaches the bottom first?

we see that the sphere is the body that arrives with the highest speed, therefore it must be the body that takes the least time,

Explanation:

Let's analyze this exercise they ask us which body reaches the bottom of the ramp first, for which we must calculate the speed of the body in the final part, use the concepts of energy conservation

starting point. Highest part of the ramp

      = U = m g h

at this point there is no kinematic energy because the body starts from rest

Final point. Lower part of the ramp

      E_{m2} = K = ½ m v² + ½ I w²

we fix the reference system at this point, as the body is rolling, it has energy of connection and rotation

energy is conserved

           E_{m1} = E_{m2}²

          mgh = ½ m v² + ½ I w²

the angular and linear variables are related

          v = wr

          w = v / r

         mg h = ½ m v² + ½ I v² / r²

          mg h = ½ m v² (1 + I / m r²)

           v² = 2gh / (1 + I / mr²)

Since all bodies start from rest, the body that has more speed at the end of the ramp, is the body that takes less time.

To find the speed we look in the tables for the moment of inertia of the bodies

circular loop    I = mr²

disk                   I = ½ m r²

sphere              I = 2/5 mr²

let's calculate the speed for each body

tie

      v² = 2g h (1+ mr² / mr²)

       v = √ (2gh) / √2

       v = 0.707 ra (2gh

disk

      v² = 2gh / (1 + ½ mr² / mr²)

      v = ra 2gh / ra 3/2

      v = 0.816 ra 2gh

sphere

      v² = 2gh / (1 + 2/5 mr₂ / mr)

      v = ra 2gh / ra 7/5

      v = 0.845 ra 2gh

we see that the sphere is the body that arrives with the highest speed, therefore it must be the body that takes the least time, since

     t = d / v

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

this is an example of "committed" time.

committed time, as contracted time, takes need over essential and available time since it is seen as profitable work. it alludes to the time allotted to keep up a home and family. at the point when a man is commuting using committed time this individual may feel that the drive is an obligation to family, for example, strolling youngsters to class or driving a mate to work. contracted and committed time clients may feel that their commute could really compare to the commute of important or leisure time clients on the grounds that their commute is profitable work. in this manner, they might be more disposed to pick a mechanized method of movement.