# Putt Physics Calculation 05: Motion of a ball rolling down a slope (1D)

Let’s talk about the physics of a ball rolling down a slope! It might sound complicated, but I promise it’s not too bad. Even though the math gets a bit trickier when we switch from a flat surface to an inclined one, the basic principles of physics still apply. Here’s a diagram of the forces at work on a ball on an inclined plane:

As you can see, the force of gravity pulling down on the ball (mg) is still the same as it would be on a flat surface. But because of the slope, we need to consider the force that’s parallel to the plane and the force that’s perpendicular to it. The parallel force is what keeps the ball moving down the slope, while the perpendicular force is what generates friction, just like we saw in the last lesson.

Since the angle of the slope affects the ball’s motion, we need to use some equations of motion to figure out what’s going on. Don’t worry if the math seems a little complicated – just bear with me. Because of the slope, gravity has a stronger effect on the ball than it would on a flat surface. This means the ball will roll downhill, and because of the angle of the slope, it will also accelerate as it goes. So we can already see that the physics of a ball rolling down a slope are different from what we saw on a flat surface.

To figure out the motion of the ball, we’ll need to use equations like:

$$F=ma$$

$$x=\frac{1}{2}at^2+v_0t$$

$$v=at+v_0$$

We’ll also need to do some trigonometry to calculate the forces acting on the ball, but the methods are the same as before. When we take the slope into account, we can calculate the straight uphill and downhill lines that the ball would follow if we were putting on the green.

Let’s start with the uphill shot! Assuming a slope angle of +2 degrees, a Stimp meter reading of 9, a friction coefficient of 0.1, and an initial ball velocity of 2m/s, the time it takes for the ball to come to a stop is:

$$t=\frac{2}{0.1*9.8}=1.96秒$$

So, the ball will come to a stop in 1.96 seconds. Using formula 2:

$$x=\frac{1}{2}(-\mu{}g)t^2+v_0t=-0.5*0.1*9.8*1.96^2+2*1.96=2.0m$$

This means that if you hit the ball with an initial velocity of 2m/s on a green with a Stimp meter reading of 9, it will roll for 2 meters. Pretty cool, huh?

I calculated the motion of a ball rolling down a slope, which is a different kind of fun than putting on a flat surface. Imagine having to consider the straight uphill and downhill lines when putting. It’s exciting to think about how you can incorporate this into your game!

By the way, you often see the term “slope angle θ°” in physics problems, but in golf, the term “gradient” is also commonly used. I personally prefer the term “angle” since it’s more familiar to me, but I wonder which term is more commonly used among golfers. If you have any opinions on which term is more understandable for most people, I’d be happy to accommodate that in my future articles.

Speaking of ball motion on a slope, once the angle exceeds 3 degrees, the ball may not stop rolling. Of course, the threshold varies depending on the condition of the green and the speed, but most greens have slopes within the range of 0° to 3° and gradients within 0% to 5%. Therefore, it’s essential to understand both slope angle and gradient when playing golf.

That concludes our calculation for the motion of a ball rolling down a one-dimensional slope.

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