Rolling charge

It is given that ball is $rolling$ and not $slipping$ . Hence there must be some force, $F$ to counter the electrostatic force, $qE$ . This force ( $F$ ):

1. provides necessary torque to $provide$ rotational motion and frictional force to $oppose$ translational motion;
2. does not allow $translational$ acceleration, $a_cm$ to exceed $rotational$ acceleration, $\alpha \cdot R$ (R = radius of sphere)

Therefore, $qE - F = m \cdot a_{cm};$ $F \cdot R = I_{cm} \cdot \alpha$ where $I_{cm} = \frac {2 \cdot M}{5 \cdot R^2}$

Using, $a_{cm} = \alpha \cdot R$

$F = \frac {2M \cdot a_{cm}}{5 \cdot R^2}$

$\Rightarrow qE = \frac {7 \cdot a_{cm}}{5}$

$\Rightarrow a_{cm} = \frac {5}{7} = 0.714 ms^{-2}$

If the ball wasn't rolling, but just slipping, applying Newton's law we would obtain $a = F/m$ , where $a$ is the acceleration of the ball, $F = q E$ the force applied on the ball by the electric field, and $m$ is the mass of the sphere.

Since the ball is indeed rolling, we can identify two different forces acting on it: the one applied by the electric field, $F$ , and a second one due to the friction between the ball surface and the plane; let's call it $F_R$ . From Newton's law we have $m a = F - F_R$ , that allows us to write $F_R = F - m a$ .

We can now write the law of the conservation of angular momentum $L$ , that writes $\frac{\Delta L}{\Delta t} = M$ , where $M$ is the torque. So we have: $M = F_R \, r = (F - ma) r$ , where $r$ is the radius of the ball, and $L = \frac{2 m r^2}{5} \omega = \frac{2 m r^2}{5} \frac{v}{r}$ , where $\frac{2 m r^2}{5}$ is the moment of intertia of a sphere and $\frac{v}{r}$ is the angular velocity of the rolling ball.

So we have $\frac{\Delta L}{\Delta t} = \frac{2 m r^2}{5} \frac{\Delta v}{\Delta t} \frac{1}{r} = \frac{2 m r^2}{5} a \frac{1}{r}$ , and equating it to the torque (and simplifying the various $r$ 's) we get $F - ma = \frac{2 m}{5} a \Rightarrow a = \frac{F}{m} \frac{5}{7}$ .

The influence of the electric force on the ball equals "F=q E =10 ⁻ ³(C) 1"(E=electric field vector) = 10⁻³N. If we put on a plane, symmetry shows that the force F acts on the center of the ball in the "+x direction".

Remembering that there is friction, which acts in the "-x direction" due to the contact with the flat surface (friction is proof that the ball rotates, instead of sliding property on its axis). This frictional force which causes a so-called "angular acceleration". Ff (frictional force) produces a torque τ=I*α about the center of mass to get the ball rolling (I=Moment of inertia, α=angular acceleration, and the magnitude of the torque is equal to τ).

Resultant force is the total force minus the dissipated = magnetic field strength (MINUS) frictional force= F-Ff.

OK. Now by parts: By Newton's 2nd Law, we get: F-Ff = m a ("a" is the linear acceleration of the center of mass) Ff=F-m a (equation we'll call "1")

The resulting torque is: Ff*r=α (which we'll call "2" and "r" is the perpendicular distance between the axis of rotation and the axis passing through the center of mass).

For the sphere I=(2/5)m*r². For a rolling object ω =v/r (r=m/radian) By this: (so gives differentiating) α=a/r.

Substituting into the equation "2" becomes: Ff r=(2/5) m r² a/r
Ff =(2/5) m a

And, from the equation "1": F-m a = (2/5) m a
F= (7/5)
m*a

The problem shows "F=10⁻³N and m=0.001 kg" 10⁻³ = (7/5) (0.001) a Finally, a= 0.714m/s².

The forces acting on the ball are the electric force $F_{e}=q E$ (applied at the center of mass of the ball), the force of friction and the normal force (both applied at the contact point between the ball and the plane). The force of friction $f$ creates a torque about the center of mass of the ball given by $\tau=fR=I \alpha$ where $\alpha$ is the angular acceleration of the ball and $I=\frac{2}{5}mR^{2}$ is the moment of inertia. Since the ball rolls without slipping we have that $a=\alpha R$ where $a$ is the acceleration of the ball's center of mass. In addition, from Newton's Second we obtain the equation $F_{e}-f= m a.$ From the above equations we find the acceleration $a= \frac{q E}{m+\frac{I}{ R^{2}}}=\frac{q E}{\frac{7}{5}m}=0.714~{\mbox{m/s}^{2}}.$