How fast is the ball going when it reaches the bottom of the hill?

Question

Nob Hill is a famous 25% grade hill in San Francisco. A ball starts rolling down the hill (starting from rest) from the top, how fast is it going when it reaches the bottom of the hill? Assume a coefficient of friction of 0.1. I roughly estimated that the hill is about 800 meters long from an image of it.

Answer 1

Hi Meg,

 

   Treat this as a conservation of energy problem.  The "input" is potential energy PE = mgh, where h is the vertical distance the ball descends, h = 0.25×800 = 200 meters in this problem.  The "output" is kinetic energy, both translational and rotational,  KE = ½mv² + ½Iω² , where I = (2/5)mr² (sphere) and v = rω; ω is the angular speed of the rolling ball  and v is the speed of the ball.  It is friction that makes the ball roll, rather than slide, down the hill.  

 

   Putting all this into an equation, where final energy equals input energy one has

 

           ½mv² + ½(2/5)mv² = mgh       (Note: the mass m will cancel (divide) out and rω has been replaced by v)  

 

You can now solve for v.      v = √ {(10/7)gh}    (Note:  this is a little smaller than if the ball slides without friction)

 

 

 

[Commentary:  Friction is normally a dissipative force,  i.e. it steals energy from your system.  One way to look at this problem is that it "steals" some translational KE and "gives" it to rotational KE.]