A boy is pushing a ring of mass 2 kg and radius 0.5 m with a stick as shown in the figure. The stick

Q: Why is the friction $f_1$ pointing downwards?

Friction opposes relative motion. The ring is rolling clockwise, meaning its left edge is moving upwards relative to the center. Since the stick only moves horizontally, the ring's left edge slides upwards against the stick. Therefore, the stick exerts a downward kinetic friction $f_1$ on the ring.

Q: Why did we use $N_1 = 2\text{ N}$ instead of the total force being $2\text{ N}$?

When a problem states 'applies a force of $2\text{ N}$' with a stick, it typically refers to the intended push (the normal force). The friction is a reactive force. If we assumed the total vector force was $2\text{ N}$, the math becomes unnecessarily complex and doesn't yield the clean integer answer expected in JEE.

Q: Why is $f_2$ pointing leftwards?

The ground friction $f_2$ is static friction that prevents the bottom of the ring from slipping. As the ring is pushed rightwards, it tends to slide rightwards at the bottom. To enforce pure rolling ($v = \omega R$), the ground exerts a leftward friction to provide the necessary clockwise torque.

Q: Can we solve this using torque about the bottom contact point?

Yes! Taking torque about the instantaneous center of rotation (the bottom point) eliminates $f_2$ and $N_2$. The torque would be $N_1 \times R - f_1 \times R = I_{\text{contact}} \alpha$. Since $I_{\text{contact}} = mR^2 + mR^2 = 2mR^2$, you get $2R - f_1 R = (2mR^2)(a/R)$, which simplifies to $2 - f_1 = 2ma$. Since $ma = 0.6$, $f_1 = 2 - 1.2 = 0.8\text{ N}$. This is a brilliant shortcut!

Question

A boy is pushing a ring of mass $2	ext{ kg}$ and radius $0.5	ext{ m}$ with a stick as shown in the figure. The stick applies a force of $2	ext{ N}$ on the ring and rolls it without slipping with an acceleration of $0.3	ext{ m/s}^2$. The coefficient of friction between the ground and the ring is large enough that rolling always occurs and the coefficient of friction between the stick and the ring is $\left(\frac{P}{10}ight)$. The value of $P$ is

Sigma Percentile · Accepted Answer

### The Illusion of the Setup

When you first read this problem, the most critical step is correctly visualizing the physical setup. The problem states that a boy is pushing a ring with a stick, applying a horizontal force of $2	ext{ N}$ to generate a horizontal acceleration. For the stick to provide a horizontal normal force $N_1$, it must be pushing against the **side** of the ring (specifically, the leftmost point at the height of the center), not resting on top of it.

If the stick were on top, the normal force would be vertical, and it couldn't possibly drive the ring forward. Thus, we establish our first anchor: $N_1 = 2	ext{ N}$ acts horizontally to the right.

### The Direction of Friction: A Common Trap

Identifying the direction of the friction force $f_1$ exerted by the stick on the ring is where many students stumble. Remember, kinetic friction always opposes *relative* motion.

As the ring accelerates to the right, it rolls clockwise. This means the leftmost point of the ring is rotating **upwards** relative to the center. The stick, however, is only moving horizontally. Therefore, the surface of the ring is sliding upwards against the stick. To oppose this upward sliding, the stick exerts a **downward** kinetic friction force $f_1$ on the ring.

Simultaneously, to maintain pure rolling, the ground exerts a static friction force $f_2$ to the left.

### The Standard Path: Newton's Laws

Let's set up our equations of motion. For translation in the horizontal direction, the net force is the forward push minus the backward ground friction:

$$ N_1 - f_2 = ma $$

Substituting the known values ($N_1 = 2	ext{ N}$, $m = 2	ext{ kg}$, $a = 0.3	ext{ m/s}^2$):

$$ 2 - f_2 = 2(0.3) \implies f_2 = 1.4	ext{ N} $$

Next, we analyze the rotational motion about the center of mass. The ground friction $f_2$ creates a clockwise torque, while the stick's friction $f_1$ creates a counter-clockwise torque. The net torque drives the clockwise angular acceleration $\alpha$:

$$ f_2 R - f_1 R = I\alpha $$

For a ring, the moment of inertia is $I = mR^2$, and the pure rolling constraint gives $\alpha = \frac{a}{R}$. Substituting these in:

$$ (f_2 - f_1)R = (mR^2)\left(\frac{a}{R}ight) = mRa $$

Dividing by $R$ yields a beautifully simple relation:

$$ f_2 - f_1 = ma $$

We already know $f_2 = 1.4	ext{ N}$ and $ma = 0.6	ext{ N}$. Plugging these in:

$$ 1.4 - f_1 = 0.6 \implies f_1 = 0.8	ext{ N} $$

### The Final Calculation

Since the stick is sliding against the ring, $f_1$ is kinetic friction, governed by the equation $f_1 = \mu N_1$.

$$ 0.8 = \mu (2) \implies \mu = 0.4 $$

The problem defines the coefficient of friction as $\frac{P}{10}$. Equating the two:

$$ \frac{P}{10} = 0.4 \implies P = 4 $$

### The Master Stroke: Instantaneous Center of Rotation

Want to solve this in half the time? Take the torque about the **bottom point of contact** (the instantaneous center of rotation). This brilliantly eliminates both $f_2$ and $N_2$ from the equation!

The normal force $N_1$ acts at a height $R$ above the bottom point, creating a clockwise torque $N_1 R$. The friction $f_1$ acts downwards at a horizontal distance $R$ from the bottom point, creating a counter-clockwise torque $f_1 R$.

Using the parallel axis theorem, the moment of inertia about the bottom point is $I_{	ext{contact}} = I_{	ext{cm}} + mR^2 = 2mR^2$.

$$ 	au_{	ext{contact}} = I_{	ext{contact}} \alpha $$

$$ N_1 R - f_1 R = (2mR^2)\left(\frac{a}{R}ight) $$

$$ N_1 - f_1 = 2ma $$

Substitute the values directly:

$$ 2 - f_1 = 2(2)(0.3) = 1.2 \implies f_1 = 0.8	ext{ N} $$

In just two lines of math, we found $f_1$, completely bypassing the ground friction. From here, $\mu = \frac{0.8}{2} = 0.4$, and $P = 4$. Elegance in physics at its finest!

Enter Numerical Value:

Visualized Solution

$System Setup$

$Free Body Diagram$

$Translational Motion$

$Solving for f_{2}$

$Rotational Motion$

$Simplifying Torque Equation$

$Solving for f_{1}$

$Coefficient of Friction$

$Final Answer$

The Sigma Insight: Dynamics of Rigid Body Rotation

The Illusion of the Setup

The Direction of Friction

The Standard Path

The Final Calculation

The Master Stroke

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } System Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Free Body Diagram

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Translational Motion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for f2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rotational Motion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying Torque Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for f1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Coefficient of Friction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer