What is the equation describing a ball motion caused by non-perpendicular force? For example in the next diagram F1 is perpendicular force but f2 is non-perpendicular force which I am asking about
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$\begingroup$ F2 will impart some linear momentum to the ball, and also cause it to rotate. Note that more than one equation applies. $\endgroup$– David WhiteApr 1, 2019 at 20:31
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$\begingroup$ You can split $F_2$ into the Normal component to the surface of the ball at the point of contact and into the Tangential component to the surface. The Normal component of $F_2$ will move the centre of mass of the ball, while the Tangential component will make the ball rotate. $\endgroup$– Ertxiem - reinstate MonicaApr 1, 2019 at 22:41
3 Answers
Its all about the centre-of-mass. I'll assume it to be in the middle of the ball.
Your $\color{blue}{\text{blue}}$ force $F_1$ points directly towards the centre-of-mass (radial). It causes only translation (it causes only a translational acceleration $a$): $$F_1=ma_y$$ The label $a_y$ indicates that it is a vertical translation.
Your $\color{red}{\text{red}}$ force $F_2$ does not point directly to the centre-of-mass. But you can split it into components: a vertical component (radial) and a horizontal component (tangential).
- The vertical (radial) component points towards the centre-of-mass and again causes translation (translational acceleration $a$). $$F_{2,rad}=ma_y$$
- The horizontal (tangential) component does not act towards the centre-of-mass and thus causes an unbalanced force on the ball spin-wise. This component causes both translation (translational acceleration $a$) and rotation (rotational acceleration $\alpha$). $$F_{2,tan}=ma_x\qquad \tau_{2,tan}=I\alpha$$ where $\tau$ is the torque created by the spin-wise unbalanced force component, and $I$ is the moment-of-inertia.
In summary, the red, tilted force $F_2$ causes rotation (counter-clockwise) as well as translation (directed Southwest). The blue $F_1$ causes only translation (directed South).
You'll have a component perpendicular to the horizontal given by Fesinθ which pushes down on the ball. It adds as a normal force. You'll also have a another component parallel to the horizontal given by Fecosθ. This force will push the ball left or right.
The verticle force, indicated by the color red, will cause both rotation and translation. If the force is applied as indicated by the color blue, it will cause translation only. The same concept applies to the horizontal force.
for simplification i take a disc instead of sphere:
force $F_1$ goes thru the center of mass thus:
$$m\,a_x=0$$ $$m\,a_y=F_1+m\,g$$ $$I_d\,\alpha=0$$ if all initial conditions are equal zero ,the disc will move toward the y direction.
for force $F_2$ we get: $$m\,a_x=F_2\,\sin(\theta)$$ $$m\,a_y=F_2\,\sin(\theta)+m\,g$$ $$I_d\,\alpha=F_2\,r=F_2\,R\,\sin(\theta)$$
if all initial conditions again equal zero and the the force $F_2$ is constant, we get $$x(t)=\frac{F_2}{m}\,\sin(\theta)\frac{t^2}{2}$$ and $$y(t)=\left(\frac{F_2}{m}\,\cos(\theta)+g\right)\frac{t^2}{2}$$
and after elimination the time $t$ we obtain
$$y(x)=a\,x$$ with $a=$ constant $$a=\cot(\theta)+\sin(\theta)\frac{m\,g}{F_2}$$ thus the disc will move a straight line
for the disc angular velocity $\omega_d$ you get:
$$\omega_d=\left(F_2\,R\,\sin(\theta)\right)\,t$$