Question

Rank the following objects by their accelerations down an incline (assume each object rolls without slipping) from least to greatest:
a. Hollow Cylinder
b. Solid Cylinder
c. Hollow Sphere
d. Solid Sphere

Answer 1

Answer:

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

Explanation:

To answer this question, let's analyze the problem. Let's use conservation of energy

Starting point. Highest point

          Em₀ = U = m g h

Final point. To get off the ramp

          Em_f = K = ½ mv² + ½ I w²

notice that we include the kinetic energy of translation and rotation

         

energy is conserved

        Em₀ = Em_f

        mgh = ½ m v² +1/2 I w²

angular and linear velocity are related

         v = w r

         w = v / r

we substitute

          mg h = ½ v² (m + I / r²)

          v² = 2 gh   $\frac{m}{m+ \frac{I}{r^2} }$

          v² = 2gh    $\frac{1}{1 + \frac{I}{m r^2} }$

this is the velocity at the bottom of the plane ,, indicate that it stops from rest, so we can use the kinematics relationship to find the acceleration in the axis ax (parallel to the plane)

         v² = v₀² + 2 a L

where L is the length of the plane

         v² = 2 a L

         a = v² / 2L

we substitute

         a = $g \ \frac{h}{L} \ \frac{1}{1+ \frac{I}{m r^2 } }$

let's use trigonometry

         sin θ = h / L

         

we substitute

         a = g sin θ   \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

the moment of inertia of each object is tabulated, let's find the acceleration of each object

a) Hollow cylinder

      I = m r²

we look for the acerleracion

      a₁ = g sin θ    $\frac{1}{1 + \frac{mr^2 }{m r^2 } }$1/1 + mr² / mr² =

      a₁ = g sin θ    ½

b) solid cylinder

       I = ½ m r²

       a₂ = g sin θ  $\frac{1}{1 + \frac{1}{2} \frac{mr^2}{mr^2} }$ = g sin θ   $\frac{1}{1+ \frac{1}{2} }$

       a₂ = g sin θ   ⅔

c) hollow sphere

     I = 2/3 m r²

     a₃ = g sin θ   $\frac{1}{1 + \frac{2}{3} }$

     a₃ = g sin θ $\frac{3}{5}$

d) solid sphere

     I = 2/5 m r²

     a₄ = g sin θ  $\frac{1 }{1 + \frac{2}{5} }$

     a₄ = g sin θ  $\frac{5}{7}$

We already have all the accelerations, to facilitate the comparison let's place the fractions with the same denominator (the greatest common denominator is 210)

a) a₁ = g sin θ ½ = g sin θ      $\frac{105}{210}$

b) a₂ = g sinθ ⅔ = g sin θ     $\frac{140}{210}$

c) a₃ = g sin θ $\frac{3}{5}$= g sin θ       $\frac{126}{210}$

d) a₄ = g sin θ $\frac{5}{7}$ = g sin θ      $\frac{150}{210}$

the order of acceleration from lower to higher is

   

     a₁ <a₃ <a₂ <a₄

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere