Golf Putt

You putt a golf ball across a green that has a uniform slope. Write equations describing the position of the golf ball as a function of time as it moves from one arbitrary location to another along the putting green. Then make a VPython animation showing the golf ball hit into the hole.

Solution

Assumptions

*The ball is a point mass with mass m.

*The green makes an angle with the horizontal.

*The ramp is steep enough that the ball is sliding.

*Static friction is the only friction acting on the ball.

*The ball is close enough to earth that g is approximately constant.

Diagrams


Analysis

According to the Force Diagram

$F_{eb}^g sin\theta - F_{rb}^s = m\ddot x$

and

$F_{rb}^n - F_{eb}^g cos\theta = m\ddot y$

These simplify to

$mgsin\theta - F_{rb}^s = m\ddot x$

and

$F_{rb}^n = F_{eb}^g cos\theta = mgcos\theta$

In order to determine $F_{rb}^s$, we can try to look at the rotational version of Newton's Second Law, $\vec{\Gamma} = \frac{d\vec{L}}{dt}$. Of the three forces acting on the ball, only the friction force exerts a torque.

$\vec{R} \times \vec{F}_{rb}^s = \frac{d\vec{L}}{dt}$

Take the magnitude of both sides

$RF_{rb}^s sin90^\circ = \frac{d}{dt}(Iw)$

$RF_{rb}^s = I\frac{dw}{dt}$

$F_{rb}^s = \frac{I}{R}\frac{dw}{dt}$

Substitute this into Newton's Second Law in the x-direction.

$mgsin\theta - \frac{I}{R}\frac{dw}{dt} = m\ddot x$

From introductory mechanics, $\frac{dw}{dt} = \alpha$ and $\alpha = \frac{\alpha_{cm}}{R}$

$mgsin\theta - \frac{I\ddot x}{R^2} = m\ddot x$

$mgsin\theta = m\ddot x + \frac{I\ddot x}{R^2} = \ddot x(m + \frac{I}{R^2}$)

$\ddot x = \frac{mgsin\theta}{m + \frac{I}{R^2}}$

For a ball that is hollow and have a thin shell, I = $\frac{2}{3}mR^2$. So

$\ddot x = \frac{3}{5}gsin\theta$

We then integrate

$\ddot x = \frac{dv}{dt} = \frac{3}{5}gsin\theta \Rightarrow \int_{v_0}^{v(t)} dv = \int_{0}^{t} \frac{3}{5}gsin\theta dt'$

$v(t) - v_0 = \frac{3}{5}gsin\theta t$

$v(t) = \frac{3}{5}gsin\theta t + v_0$

$v(t) = \frac{dx}{dt} = \frac{3}{5}gsin\theta t + v_0 \Rightarrow \int_{x_0}^{x(t)} dx = \int_{0}^{t} \frac{3}{5}gsin\theta t' + v_0$

$x(t) - x_0 = \frac{3}{5}gsin\theta t^2 + v_0 t$

$x(t) = \frac{3}{5}gsin\theta t^2 + v_0 t + x_0$

We can use these to find the corresponding rotational motion.

$\alpha = \frac{\ddot x}{R} = \frac{\frac{3}{5}gsin\theta}{R}$

$\alpha = \frac{wt}{dt} = \frac{\frac{3}{5}gsin\theta}{R} \Rightarrow \int_{w_0}^{w(t)} dw = \int_{0}^{t} \frac{\frac{3}{5}gsin\theta}{R}dt'$

$w(t) - w_0 = \frac{\frac{3}{5}gsin\theta}{R}t$

$w(t) = \frac{\frac{3}{5}gsin\theta}{R}t + w_0$

$\phi(t) - \phi_0 = \frac{gsin\theta}{5R}t^2 + w_0t$

$\phi(t) = \frac{gsin\theta}{5R}t^2 + w_0t + \phi_0$

In the limit that $\theta \to 0$, we would expect that there is no acceleration at all.

Interpretation

The linear acceleration of the golf ball is much smaller than that of an object that is sliding without friction, which makes sense. Realistically our result would likely break down at a high enough angle because the golf ball would begin to slide instead of roll. This system would also fall apart if the initial force is 0. Since the ball can't physicslly move on it's own and we know that the only other two forces acting on the ball is gravity and static friction. Static friction would also be zero because it counters the normal force which is 0. So the ball would just not go anywhere at all.

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