# Problem of the week #52 - March 25th, 2013

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#### Jameson

Staff member
** Today is a big milestone for our POTW contest! We've been doing this for 1 year now without missing a single week. We hope you all continue to enjoy these problems and also that this next year we can find more exposure for these problems so students of all levels can push themselves to apply what they already know in challenging ways. ** [HR][/HR]

This week's problem is one of the most challenging we've had for this level and it requires some knowledge of physics concepts. Thank you to MarkFL for submitting this problem for us to use!

A cue strikes a cue ball and delivers a horizontal impulse in such a way that the ball rolls without slipping as it starts to move. At what height above the ball's center (in terms of radius of the ball) was the blow struck?

In order to solve this you will need to know the following:

1) Moment of inertia of a solid sphere: $\displaystyle I=\frac{2}{5}MR^2$

2) Parallel-axis theorem: $\displaystyle I=I_{\text{CM}}+MD^2$

3) Net torque on the ball: $\displaystyle \sum\vec{\tau}_P=F_C(R+x)=I\alpha$.

Here is a hint that explains the situation in terms of approaching the solution:
The first condition of equilibrium yields no useful information because we know nothing of the nature of the friction between the table's surface and the cue ball. We can, however, use the fact that the ball is not in rotational equilibrium and analyze the torque resulting from the horizontal impulse. Since we know nothing about the force of static friction which keeps the ball from slipping, we will choose for the axis of rotation the contact point $P$ between ball and surface. Here the moment arm for the force of static friction is zero. Let $x$ be the requested height above the ball's center, $R$ be the ball's radius, $M$ be the ball's mass, $F_C$ be the average force on the ball, and $a$ be the acceleration of the ball.

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#### Jameson

Staff member
Congratulations to the following members for their correct solutions:

since at the beginning the ball is at rest,
If the impulse is $P$ and after the impulse if the angular velocity of the ball is $\omega$
\begin{align*}Fr&=I\alpha \\ \int Fr\;dt&=\int I \alpha \;dt\\ \left(\int F\;dt\right) \;r&= I \left(\int \alpha \;dt\right)\\ Pr&=I(\omega - 0)\\ Pr&=I\omega \end{align*}
applying $Pr=I\omega$, to the point of contact of the ball and the surface $P$

$$P(R+x)=I_P\omega$$
($R$ is the radius of the sphere and $x$ is the vertical distance of the contact of the cue from the center)

From Parallel axis theorem,

$$I_p=I_0+mR^2$$
when $I_0=\frac{2}{5}mR^2$
$$I_p=\frac 25mR^2+mR^2=\frac 75mR^2$$

$$P(R+x)=\frac 75mR^2 \omega \qquad (1)$$

for globe to not to slip,
$$\omega R = v \qquad (2)$$ where $v$ is the velocity of the sphere just after the contact of the sphere

applying $\Delta I=m\Delta v$ to the ball
$$P=m(v-0)\implies P=mv \qquad (3)$$

from (1),(2),(3)

\begin{align*} mv(R+x)&=\frac 75mR^2\frac vR\\ R+x&=\frac 75 R\\ x&=\frac 2 5 R \end{align*}

Therefore the cue should contact the ball at a length of 2/5 times the radius of the ball above the center for the ball to avoid slipping

(from MarkFL):
The torque on the ball is given by:

$$\displaystyle \sum\vec{\tau}_P=F_C(R+x)=I\alpha$$

Newton's 2nd law of motion gives us:

$$\displaystyle F_C=Ma$$

The moment of inertia, along with the parallel-axis theorem gives is the moment of inertia for the ball:

$$\displaystyle I=\frac{2}{5}MR^2+MR^2=\frac{7}{5}MR^2$$

The relationship between angular acceleration and linear acceleration is:

$$\displaystyle \alpha=\frac{a}{R}$$

and so we find:

$$\displaystyle F_C(R+x)=\frac{7}{5}MR^2\frac{a}{R}=\frac{7}{5}F_CR\,\therefore\,x=\frac{2}{5}R$$

Thus, the cue strikes the cue ball at a height of 2/5 of the radius of the ball above the center.

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