Question

A uniform slender rod of length $l$ and mass $m$ is rigidly to a circular hoop of radius $l$ as shown. The mass of the hoop is negligible. If the rod and hoop are released from rest on a fixed horizontal surface in the position illustrated, determine the initial value of the friction force F. if friction is sufficient to prevent slipping.

• A. $\frac{8mg}{3}$
• B. $\frac{2mg}{7}$
• C. $\frac{3mg}{5}$
• D. $\frac{3mg}{8}$

Answer 1

Answer: (D)

$\displaystyle \frac{3mg}{8}$

Answer 2

We solve the problem by “rotating‐about the instantaneous axis” (point of contact P). In our analysis we assume that the rod is initially horizontal. (This is what the diagram implies.) Then:

1. Choose coordinates so that the circular hoop (radius l) touches the floor at point P = (0,0). Its center is at O = (0,l) and the rod is rigidly attached so that, being horizontal, its center-of-mass (of the uniform rod of length l) is at
   G = O + (l/2, 0) = (l/2, l).

2. The only external force with a moment about P is the weight mg acting vertically downward at G. The torque about P is

   $$\tau = mg \times (\mbox{horizontal distance from }P\mbox{ to line of action of }mg) = mg \times \frac{l}{2} = \frac{mgl}{2}\,.$$

3. The assembly rotates about P; we compute the moment of inertia by the parallel‐axis theorem. The rod’s moment of inertia about its center for rotation about an axis perpendicular to it is

   $$I_{cm}=\frac{1}{12}ml^2\,.$$

   The distance from G to P is

   $$d=\sqrt{\left(\frac{l}{2}\right)^2+l^2}=\frac{l\sqrt{5}}{2}\,.$$

   Thus,

   $$I_P=I_{cm}+m\,d^2=\frac{1}{12}ml^2+m\left(\frac{l\sqrt{5}}{2}\right)^2=\frac{1}{12}ml^2+ \frac{5ml^2}{4}=\frac{ml^2}{12}\left(1+15\right)=\frac{16ml^2}{12}=\frac{4ml^2}{3}\,.$$

4. The angular acceleration about P is

   $$\alpha=\frac{\tau}{I_P}=\frac{(mgl/2)}{(4ml^2/3)}=\frac{3g}{8l}\,.$$

5. Since the whole body rotates about P, each point moves in a circle centered at P. The center-of-mass G is at a distance $d=\frac{l\sqrt{5}}{2}$ and its tangential acceleration is $d\,\alpha$. However, using geometry one may show that the horizontal component of this acceleration equals exactly

   $$a_x=l\,\alpha=\frac{3g}{8}\,.$$

   (Details: The tangent to the circle at G is perpendicular to $\vec{PG}=(l/2,\,l)$; working out its components gives $a_x=\alpha\,(l\sqrt{5}/2)\,(2/\sqrt{5})=l\alpha$.)

6. The only horizontal force on the body is the friction force $F$. By Newton’s second law,

   $$F = m a_x = m\left(\frac{3g}{8}\right)=\frac{3mg}{8}\,.$$

Thus the initial friction force is $\displaystyle \frac{3mg}{8}$. This corresponds to option (D).

Explanation (minimal):

• Choose P as instantaneous axis (contact point).
• The weight mg at the rod’s COM (at $(l/2,l)$) gives a torque $\tau=\frac{mgl}{2}$.
• Using parallel‐axis theorem, the moment of inertia about P is $I_P=\frac{4ml^2}{3}$.
• Therefore, $\alpha=\frac{3g}{8l}$.
• The horizontal acceleration of the COM is $a_x=l\alpha=\frac{3g}{8}$.
• So, $F=ma_x=\frac{3mg}{8}$.