91Ó°ÊÓ

The free-body diagram of the ball that is rolling down the incline without sleeping is shown below:

From the diagram, we observed that there is a weight acting on the bowling ball, there is force of friction which acts in the direction that is opposite to the direction of motion.

Therefore, it is directed up the incline (it slows the ball down) and normal reaction force.

Another reason is torque for which the friction force is directed up the incline.

Form the diagram, we see that as the ball is moving up the incline, therefore the linear velocity of the centre of the mass will be directed up the incline, and

therefore, the angular velocity will be directed clockwise, and since linear acceleration is directed down the incline, angular acceleration will be directed counter clockwise.

Now, the net torque of the center of mass is,

$\mathrm{Ï„}=I\mathrm{Î\pm }$, where,

I is moment of inertia, $\mathrm{Î\pm }$is angular acceleration.

Since $I>0$, the torque will be in the same direction as that of angular acceleration, which is counter clockwise.

The torque exerted by the force of friction is,

$\mathrm{Ï„}=F_{fr}R$, where,

$F_{fr}$is the force of friction and R is the radius.

Hence, from these above two torque equations, we can conclude that the force of friction has to be directed up the incline in order to produce the torque that is directed counter clockwise.

We have the torque,

$\mathrm{Ï„}=I\mathrm{Î\pm }$.

The moment of inertia is $I=\frac{2}{5}m{R}^2$, substituting in the above equation, we get,

$\mathrm{Ï„}=\frac{1}{2}m{R}^2\mathrm{Î\pm }$.

Now, applying Newton’s Second Law, in this case, we assume that the net force will be the combination of component of weight and force of friction:

$ma=mg\sin \mathrm{β}+F_{fr}$.

The angular acceleration is,

$\mathrm{Î\pm }=\frac{a}{R}$

We will combine the equation that determines the torque and angular acceleration in order to represent force of friction.

Thus,$F_{fr}=\frac{2}{5}ma$.

Next, we will combine this equation that determines force of friction with the equation that determines net force to represent acceleration.

$a=\frac{5}{7}g\sin \mathrm{β}$.

Hence, the acceleration of the center of mass is, $a=\frac{5}{7}g\sin \mathrm{β}$.

Here, we will assume that

$F_{fr}≤{\mathrm{μ}}_{s}N$

Where, ${\mathrm{μ}}_s$is coefficient of static friction and N is normal reaction force.

Let us apply the Newton’s Second Law to determine the net force in the vertical direction, in which it will be the combination of component of weight and normal reaction force which is equal to 0.

Thus,

$$\begin{gathered} N-mg\cos \mathrm{β}=0 \\ ⇒N=mg\cos \mathrm{β} \end{gathered}$$

Therefore,

$$\begin{gathered} F_{fr}≤{\mathrm{μ}}_{s}mg\cos \mathrm{β} \\ ⇒\frac{2}{5}ma≤{\mathrm{μ}}_{s}mg\cos \mathrm{β} \\ ⇒\frac{2}{5}a≤{\mathrm{μ}}_{s}g\cos \mathrm{β} \\ ⇒\frac{2}{5}\frac{5}{7}g\sin \mathrm{β}≤{\mathrm{μ}}_{s}g\cos \mathrm{β} \\ ⇒\frac{2}{7}\tan \mathrm{β}≤{\mathrm{μ}}_s \end{gathered}$$

Hence, the minimum coefficient of static friction that prevent slipping is, $\frac{2}{7}\tan \mathrm{β}≤{\mathrm{μ}}_s$.