Question

You want to race a hoop, I_{hoop} = MR^2I hoop =MR 2 , a solid sphere, I_{solid sphere} = \frac{2}{5}MR^2I solidsphere = 5 2 MR 2 and a solid cylinder, I_{solid cylinder} = \frac{1}{2}MR^2I solidcylinder = 2 1 MR 2 down a ramp. Each object has the same mass and radius. Predict which object will win the race.A. The hoop will win.B. The solid sphere will win.C. The solid cylinder will win.D. All three objects will tie.E. It's impossible to tell

Answer 1

Answer:

The fastest object is the sphere, so it is the winner

Explanation:

To know which object will arrive faster down, let's look for the velocity of the center of mass of each object. Let's use the concept of mechanical energy

Highest point

      Em₀ = U = mg y

       

Lowest point

     $Em_{f}$= K = $K_{rot}$ + $K_{cm}$ = ½ I w² + ½ m $v_{cm}$²

Angular velocity is related to linear velocity.

       v = w r

       w = v / r

        $Em_{f}$ = ½ I $v_{cm}$²/r² + ½ m $v_{cm}$²

        $Em_{f}$ = ½ (I / r² + m) $v_{cm}$²

Energy is conserved

      Em₀ =  $Em_{f}$

      mg y = ½ (I / r² + m) $v_{cm}$²

      $v_{cm}$ = √2 g y / (I / mr² +1)

With this expression we can know which object arrives as a higher speed, therefore invests less time and is the winner. Let's calculate the speed of the center of mass of each

Ring

       I = m r²

      $v_{cm}$ = √ (2 g y / (m r² / mr² + 1))

      $v_{cm}$ = √ (2gy 1/2)

      $v_{cm}$ = (√ 2gy) 0.707

Solid sphere

      I = 2/5 m r²

     $v_{cm}$ = √ (2gy / (2/5 m r² / mr² + 1)

     $v_{cm}$ = √ (2gy / (7/5))

     $v_{cm}$ = √ (2gy 5/7)

     $v_{cm}$ = (√ 2gy) 0.845

Cylinder

      I = ½ m r²

      $v_{cm}$ = √ (2gy / ½ mr² / mr² + 1)

      $v_{cm}$ = √ (2gy / (3/2))

      $v_{cm}$ = √ (2g y 2/3)

      $v_{cm}$ = (√ 2gy) 0.816

The fastest object is the sphere, so it is the winner when descending the ramp