This problem involves a ring being pushed by a stick while rolling without slipping on the ground. We are to find the coefficient of friction between the stick and the ring, given as P/10. Let's analyze the forces and torques acting on the ring.

Step 1: Identify the Forces

The ring has a mass $m=2\mathrm{kg}$ and radius $R=0.5\mathrm{m}$. The stick applies a force $F=2\mathrm{N}$ at some point (the figure shows it applied tangentially, which we will assume). The ring accelerates with $a=0.3{\mathrm{m}}^{/{\mathrm{s}}^2}$.

Key forces:

• The applied force from the stick, $F=2\mathrm{N}$, acting tangentially to the ring.
• Friction force between the ground and the ring, $f_g$, which acts opposite to the direction of motion to prevent slipping.
• Friction force between the stick and the ring, $f_s$. Since the stick is pushing the ring, this friction acts along the direction of motion to help rolling.
• Normal forces and weight act vertically, but since motion is horizontal, we focus on horizontal forces.

Step 2: Equations of Motion

For translational motion (Newton's second law):

$F+f_s-f_g=ma$

Substituting known values:

$2+f_s-f_g=2\times 0.3=0.6$

So,

$2+f_s-f_g=0.6$

Rearranging:

$f_s-f_g=-1.4$

Equation (1): $f_g-f_s=1.4$

For rotational motion (torque about the center): The friction force from the ground $f_g$ provides a clockwise torque, and the friction from the stick $f_s$ and the applied force $F$ provide counterclockwise torques. Assuming the stick applies force tangentially:

$FR+f_{s}R-f_{g}R=I\alpha$

For a ring, moment of inertia about center is $I=m{R}^2$. For rolling without slipping, $\alpha =\frac{a}{R}$.

Substitute:

$FR+f_{s}R-f_{g}R=m{R}^2·\frac{a}{R}=mRa$

Divide both sides by R:

$F+f_s-f_g=ma$

But this is the same as the translational equation we already used. So, we need to consider torque about another point to get an independent equation. Let's consider torque about the point of contact with the ground.

Step 3: Torque About Point of Contact

Taking torque about the point of contact with the ground (which is instantaneously at rest), the only force producing torque is the applied force $F$ and the stick friction $f_s$, both at the top of the ring. Their lever arm is the diameter, 2R.

Net torque = $(F+f_s)·2R$

This torque equals the moment of inertia about the point of contact times angular acceleration. For a ring, using parallel axis theorem, $I_{\mathrm{contact}}=I+m{R}^2=m{R}^2+m{R}^2=2m{R}^2$.

So,

$(F+f_s)·2R=2m{R}^2\alpha$

But $\alpha =\frac{a}{R}$, so:

$2R(F+f_s)=2m{R}^2·\frac{a}{R}=2mRa$

Divide both sides by 2R:

$F+f_s=ma$

Substitute known values:

$2+f_s=2\times 0.3=0.6$

So,

$f_s=0.6-2=-1.4\mathrm{N}$

The negative sign indicates that our assumed direction for $f_s$ (same as F) is wrong. Actually, $f_s$ acts opposite to F, with magnitude 1.4 N.

So, $f_s=1.4\mathrm{N}$ (acting backwards).

Step 4: Find Coefficient of Friction Between Stick and Ring

The friction force $f_s$ is related to the normal force. The stick applies a normal force $N$ (which is the applied force 2N, since it is perpendicular to the ring). So,

$f_s=\mu N=\mu ·2$

We found $f_s=1.4\mathrm{N}$, so:

$2\mu =1.4$

Thus,

$\mu =\frac{1.4}{2}=0.7$

But the problem states the coefficient is P/10, so:

$\frac{P}{10}=0.7$

Therefore, $P=7$.

Final Answer: P = 7

Related Topics and Formulae

Rolling Without Slipping: Combines translation and rotation such that the point of contact has zero velocity relative to the surface. Condition: $v=\omega R$ and $a=\alpha R$.

Torque About Instantaneous Point of Contact: Useful for solving rolling problems, as it simplifies equations by eliminating the friction force at the contact point.

Friction: $f_s\le \mu N$, where $\mu$ is the coefficient of friction and N is the normal force.