Question

A solid ball is released from rest and slide down a hill-side that slopes downward$\mathbf{65}\mathbf{.}\mathbf{0}\mathbf{°}$from the horizontal. (a) What minimum value must the coefficient of static friction between the hill and ball surface have for no slipping to occur? (b) Would the coefficient of friction calculated in part (a) be sufficient to prevent a hollow ball (such as soccer ball) from slipping? Justify your answer. (c) In part (a), why did we use the coefficient of static friction and not the coefficient of kinetic friction?

Short answer

(a) The coefficient of static friction is, ${\mu }_s=0.613$.

(b) The ball will be slipping for any value of coefficient of static friction between these two values.

(c) There was no relative motion between the solid sphere and the hill.

Step-by-step solution

(a)To find the minimum value of the coefficient of static friction

In this first part, we will use Newton’s Second Law, that shows the net force is the product of the acceleration of the center of mass and its mass.

From the diagram shown, we see that there are two forces:

Normal reaction force, and vertical component of weight.

Therefore, the Newton’s Second Law will be,

$F_{nety}=n-mg\cos \theta \cdots \cdots \left(1\right)$

Next, forces acting in horizontal direction, there are two forces: force of friction and horizontal component of weight.

Therefore, we have,

$ma=mg\sin \theta -F \cdots \cdots \left(2\right)$, where,

$F={\mu }_{s}mg\cos \theta$, where ${\mu }_s$is the coefficient of static friction.

Thus, equation $\left(2\right)$becomes,

$$\begin{gathered} ma=mg\sin \theta -{\mu }_{s}mg\cos \theta \\ \Rightarrow a=g\left(\sin \theta -{\mu }_s\cos \theta \right) \end{gathered}$$

Now, the torque exerted by the force of friction is,

${\tau }_F={\mu }_{s}mg\cos \theta R$.

Also, the torque can be expressed as,

$\tau =I\alpha$,

where $\alpha =\frac{a}{R}$, is an angular acceleration and R is the radius, $I=\frac{2}{5}m{R}^2$is the moment of inertia of the center of mass.

Thus, equating both the torques, we get,

$$\begin{gathered} {\mu }_{s}mg\cos \theta R=I\alpha \\ \Rightarrow {\mu }_{s}mg\cos \theta R=\frac{2}{5}m{R}^2\alpha \\ \Rightarrow {\mu }_{s}g\cos \theta =\frac{2}{5}a \cdots \cdots \left(2\right) \end{gathered}$$

Since angular acceleration from Newton’s second Law can be expressed as:

$a=\frac{5}{7}g\sin \theta$.

Putting this value in$\left(2\right)$, we get coefficient of friction as:

$$\begin{gathered} \Rightarrow {\mu }_{s}g\cos \theta =\frac{2}{5}·\left(\frac{5}{7}g\sin \theta \right) \\ \Rightarrow {\mu }_s=\frac{2}{7}\tan \theta \end{gathered}$$

Now, substituting values, we get,

$$\begin{gathered} {\mu }_s=\frac{2}{7}\tan 65{}^{\circ } \\ \Rightarrow {\mu }_s=0.613 \end{gathered}$$

Hence, the coefficient of static friction is, ${\mu }_s=0.613$.

(b)To find the coefficient of friction is sufficient to prevent the ball from slipping

We have the moment of inertia of the hollow sphere as:

$I=\frac{2}{3}m{R}^2$.

Therefore, acceleration is,

$a=\frac{3}{2}g\sin \theta$.

From torque balance, the coefficient of friction will be expressed (with new moment of inertia) as:

${\mu }_s=\frac{2}{5}\tan \theta$.

Now, substituting values, we get,

localid="1667970152706" $$\begin{gathered} {\mu }_s=\frac{2}{5}\tan 65{}^{\circ } \\ {\mu }_s=0.858 \end{gathered}$$

We observed that the coefficient of static friction of hollow sphere is greater than the coefficient of static friction of the solid sphere.

Hence, the ball will be slipping for any value of coefficient of static friction between these two values.

(c)To state why we used coefficient of static friction in part (a)

Since there was no relative motion between the solid sphere and the hill, we used coefficient of static friction and not coefficient of kinetic friction.

If there was any motion, we would be using coefficient of kinetic friction.