# How many times will a ball spin if I drop it of a height $h$? [closed]

I'm currently doing an extended project qualification and my topic is to form an equation which maximizes the probability of the ball landing on a point marked with a red pen on the ball.

Assuming that the ball is exactly spherical, with a diameter of 'd'. The ball is rolled of a curved downwards surface to make friction negligible . The ball moves down a vertical distance of height H_t and is in contact with the surface for a distance of twice the circumference of the ball. It starts rotating when the red dot touches the surface and rolls off when the red dot touches the surface. So as sum of energy is conserved KE = GPE V_final when the ball leaves the surface is V_final = sqrt(2gH_t). The ball then undergo projectile motion with horizontal speed of sqrt(2gH_t) and moves a horizontal distance of R. And it moves down a vertical distance of H_f as it reaches the ground. Using the idea of Torque = rotational inertia x angular acceleration is it possible to determine how many times the ball will spin before it touches the ground. or for a time of T seconds which can be calculated by the verticle distance.

I need some recommendations of physics concepts I should look at for example, newton's second law of rotation, and etc to be able to determine this. Also, i want to apply air resistance to the system.

Given area of the sphere incontact with the air is, pi x d^2 , Drag coefficient is; C_d Density of air is ; Rho

My plan is once i get torque = rotational inertia x angular acceleration, I can integrate both sides twice to get angular displacement setting boundary conditions

## closed as unclear what you're asking by garyp, Jon Custer, user36790, David Z♦Oct 27 '16 at 18:28

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• I'm voting to close this question as off-topic because there is no evidence of prior effort. – garyp Oct 27 '16 at 10:50
• Please tell us what you've done so far, or what ideas you have. Please also state the conditions in more detail : Is the ball dropped vertically? Do you spin the ball as you release it? Do you roll the ball off the side of a table? (We need to know distance the ball goes horizontally from the table.) What is the diameter of the ball? What is the orientation of the red spot when the ball is dropped? Is it on top or bottom? – sammy gerbil Oct 27 '16 at 15:44

First you need to work out the rotational speed $\omega$ of the ball, which will be constant as it falls. (There is no torque acting on the ball while it is in flight, if drag is neglected.) Assuming the ball is rolling and not slipping at the point of launch, $\omega$ is related to the linear speed $v$ of the CM at the point of launch by $v=\omega d$. You also need the time of flight $T$.
Both $v$ and $T$ can be worked out from the height $H$ and range $R$ of the projectile.
If you do not know the range $R$, you can calculate $v$ and $\omega$ from the potential energy lost as the ball rolls down the slope. Assuming the ball does not slip as it rolls down the slope, you can use conservation of energy - as you have attempted to do, but you have neglected rotational KE. So you should have
$mgH_1=\frac12mv^2+\frac12J\omega^2$
where $J=\frac25md^2$ is the moment of inertia about the centre. However, this calculation is not as reliable as using the range $R$, because it assumes there is no slipping at any point on the slope, whereas using $R$ only assumes no slipping at the launch point.
Finally, to land on the red spot the ball must complete a whole number $n$ of revolutions in time of flight $T$ : ie $n=(\omega/2\pi)T$.