A man pushes a cylinder of mass m1 with the help of a plank of mass m2 as shown. There is no slippin

Q: Why is the acceleration of the plank twice the acceleration of the cylinder's center of mass?

Because there is no slipping at the ground, the bottom of the cylinder is instantaneously at rest, making $a_2 = R\alpha$. The top of the cylinder, which is in contact with the plank, has an acceleration of $a_2 + R\alpha = 2a_2$. Since the plank doesn't slip on the cylinder, it must share this acceleration, so $a_1 = 2a_2$.

Q: How do we know the direction of the frictional forces?

We can assume any arbitrary direction initially. If our final calculated value is positive, our assumed direction was correct. If it's negative, the actual direction is opposite. Here, assuming both $f_1$ and $f_2$ act forwards on the cylinder yields positive results.

Q: Why doesn't the applied force $F$ create a torque on the cylinder?

The force $F$ is applied to the plank, not directly to the cylinder. The plank interacts with the cylinder only through the frictional force $f_1$ at their contact surface. It's this friction $f_1$ that exerts a torque on the cylinder.

Q: What is the moment of inertia of the cylinder?

For a solid uniform cylinder rotating about its central axis, the moment of inertia is $I = \frac{1}{2}m_1 R^2$. This is a standard result derived from integrating the mass distribution.

Q: Can we solve this using energy methods?

While energy methods are powerful, they are less direct here because we need to find the internal frictional forces and accelerations at a specific instant. Newton's laws and torque equations provide a more straightforward path to these specific quantities.

Question

A man pushes a cylinder of mass $m_1$ with the help of a plank of mass $m_2$ as shown. There is no slipping at any contact. The horizontal component of the force applied by the man is $F$. Find
(a) the accelerations of the plank and the centre of mass of the cylinder and
(b) the magnitudes and directions of frictional forces at contact points.

Sigma Percentile · Accepted Answer

### The Setup and the Challenge

Imagine a plank resting on a cylinder, which in turn rests on the ground. You push the plank with a force $F$. The catch? There is absolutely no slipping anywhere. This means the plank grips the cylinder perfectly, and the cylinder grips the ground perfectly.

Our goal is to find the accelerations of both the plank and the cylinder, and to uncover the hidden frictional forces that make this pure rolling possible.

### Unlocking the Kinematic Constraints

Before we write any force equations, we must understand how the motions of the plank and the cylinder are locked together. Let's define our variables: let $a_1$ be the acceleration of the plank, $a_2$ be the acceleration of the cylinder's center of mass, and $\alpha$ be the cylinder's angular acceleration.

Because the cylinder rolls without slipping on the ground, the velocity and acceleration of its bottommost point must be zero. This gives us our first crucial relationship: $a_2 = R\alpha$.

Similarly, the top of the cylinder must move exactly with the plank to avoid slipping. The acceleration of the top point of the cylinder is the sum of its translational and rotational accelerations: $a_2 + R\alpha$. Since this must equal the plank's acceleration, we have $a_1 = a_2 + R\alpha$.

Substituting our first relationship into the second, we find a beautiful constraint: **$a_1 = 2a_2$**. The plank always accelerates twice as fast as the cylinder's center!

### The Dance of Forces: Free Body Diagrams

Now, let's break the system apart and look at the forces.

For the plank, the applied force $F$ pushes it forward. But the cylinder resists this motion through friction. Let's call this friction $f_1$, acting backwards on the plank. Newton's second law tells us:

$F - f_1 = m_2 a_1$

By Newton's third law, if the cylinder pulls the plank backwards with $f_1$, the plank pulls the cylinder forwards with $f_1$.

Now, what about the friction between the cylinder and the ground? Let's call it $f_2$. We don't know its direction yet, so let's assume it acts forwards. If our final answer is positive, we guessed right! The translational equation for the cylinder is:

$f_1 + f_2 = m_1 a_2$

### The Torque Equation

The cylinder isn't just translating; it's rotating. The frictional forces $f_1$ and $f_2$ create torques about the center of mass.

The top friction $f_1$ creates a clockwise torque of $f_1 R$. The bottom friction $f_2$ (which we assumed acts forwards) creates an anti-clockwise torque of $-f_2 R$. The net torque drives the angular acceleration:

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

We know the moment of inertia of a solid cylinder is $I = \frac{1}{2}m_1 R^2$, and from our constraints, $\alpha = \frac{a_2}{R}$. Substituting these in, the radius $R$ elegantly cancels out:

$f_1 - f_2 = \frac{1}{2}m_1 a_2$

### Bringing It All Together

We now have a neat system of equations. Let's solve for the frictions first. We have:

$f_1 + f_2 = m_1 a_2$

$f_1 - f_2 = \frac{1}{2}m_1 a_2$

Adding these equations eliminates $f_2$, giving us $2f_1 = \frac{3}{2}m_1 a_2$, which means **$f_1 = \frac{3}{4}m_1 a_2$**.

Subtracting them eliminates $f_1$, giving us $2f_2 = \frac{1}{2}m_1 a_2$, which means **$f_2 = \frac{1}{4}m_1 a_2$**.

Now, let's find the acceleration. We substitute $f_1$ and our constraint $a_1 = 2a_2$ back into the plank's equation:

$F - \frac{3}{4}m_1 a_2 = m_2 (2a_2)$

Rearranging to solve for $a_2$:

$F = \left(\frac{3}{4}m_1 + 2m_2\right) a_2$

**$a_2 = \frac{4F}{3m_1 + 8m_2}$**

This is the acceleration of the cylinder. The plank's acceleration is simply twice this value:

**$a_1 = \frac{8F}{3m_1 + 8m_2}$**

### The Final Reveal

Finally, we can find the exact values of the frictional forces by plugging $a_2$ back into our expressions for $f_1$ and $f_2$:

**$f_1 = \frac{3m_1 F}{3m_1 + 8m_2}$**

**$f_2 = \frac{m_1 F}{3m_1 + 8m_2}$**

Notice that both $f_1$ and $f_2$ came out positive. This confirms that our initial assumptions about their directions were absolutely correct: $f_1$ acts forwards on the cylinder (and backwards on the plank), and $f_2$ acts forwards on the cylinder. The physics works out perfectly!

Visualized Solution

System Setup

Kinematic Constraints

Equation of Motion for Plank

Equation of Motion for Cylinder (Translation)

Equation of Motion for Cylinder (Rotation)

Solving for Frictions

Frictions in terms of $a_{2}$

Finding Acceleration $a_{2}$

Finding Acceleration $a_{1}$

Final Frictional Forces

The Sigma Insight: Rolling Motion

The Setup and the Challenge

Unlocking the Kinematic Constraints

The Dance of Forces

The Torque Equation

Bringing It All Together

The Final Reveal

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