Calculate Coefficient of Friction Between Stick and Rolling Ring

To find the value of $P$, we analyze the translational and rotational motion of the ring.

Given Data:

• Mass of the ring, $m = 2\text{ kg}$
• Radius of the ring, $R = 0.5\text{ m}$
• Force applied by the stick (normal force on the ring), $N = 2\text{ N}$
• Linear acceleration of the ring, $a = 0.3\text{ m/s}^2$
• Coefficient of friction between the stick and the ring, $\mu = \frac{P}{10}$

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1. Condition for Pure Rolling

Since the ring rolls without slipping on the ground, the relation between the linear acceleration $a$ and the angular acceleration $\alpha$ is:
$a = R\alpha \implies \alpha = \frac{a}{R}$

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2. Equations of Motion

A. Translational Motion (Horizontal Direction)

Assuming the stick pushes the ring to the left with normal force $N = 2\text{ N}$, the ring accelerates to the left. The static friction force from the ground, $f_g$, opposes this motion and acts towards the right.

Using Newton's second law for horizontal motion:
$N - f_g = m a$

Substitute the known values:
$2 - f_g = 2 \times 0.3$
$2 - f_g = 0.6 \implies f_g = 1.4\text{ N}$

B. Rotational Motion (Torque about Center of Mass)

Taking the counterclockwise rotation as positive:

• The moment of inertia of a ring about its central axis is $I = m R^2$.
• The friction force from the ground, $f_g$, acts to the right at the bottom point, creating a counterclockwise torque:
  $\tau_g = f_g R$
• The friction force from the stick, $f_s$, acts at the rightmost contact point of the ring. To oppose the upward relative velocity of the contact point on the rotating ring, $f_s$ acts downwards, creating a clockwise torque:
  $\tau_s = -f_s R$

Using Newton's second law for rotation:
$\tau_{\text{net}} = I \alpha$
$f_g R - f_s R = (m R^2) \left(\frac{a}{R}\right)$

Dividing both sides by $R$:
$f_g - f_s = m a$

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3. Calculation of Friction Force $f_s$ and $P$

Substitute $f_g = 1.4\text{ N}$, $m = 2\text{ kg}$, and $a = 0.3\text{ m/s}^2$:
$1.4 - f_s = 2 \times 0.3$
$1.4 - f_s = 0.6 \implies f_s = 0.8\text{ N}$

The friction force exerted by the stick on the ring is given by:
$f_s = \mu N$

Substitute $f_s = 0.8\text{ N}$ and $N = 2\text{ N}$:
$0.8 = \mu \times 2 \implies \mu = 0.4$

Given that $\mu = \frac{P}{10}$:
$\frac{P}{10} = 0.4 \implies P = 4$