Two thin circular discs of mass 2 kg and radius 10 cm each are joined by a rigid massless rod of len

Q: Why is the friction force acting in the forward direction?

When the truck accelerates forward, the disc's inertia makes it want to slide backward relative to the truck. Friction opposes this relative motion, so it acts in the forward direction to pull the disc along with the truck.

Q: Why is the acceleration of the contact point $a_0 + R\alpha$ and not $a_0 - R\alpha$?

The forward friction force creates a counter-clockwise torque about the center of mass. A counter-clockwise angular acceleration means the bottom point of the disc accelerates forward relative to the center of mass. Since the center of mass is also accelerating forward, the two components add up.

Q: How do we know the rod is along the Y-axis?

The problem states the truck moves along the X-axis and the Z-axis is vertically upwards. It also says the axis of the object is horizontal and perpendicular to the direction of motion. The only horizontal direction perpendicular to X is the Y-axis.

Q: Why do we calculate the torque about the center of mass $O$?

The problem explicitly asks for the frictional torque acting on each disc about the center of mass $O$ of the object. The center of mass is the midpoint of the rod connecting the two discs.

Q: Does the mass of the rod affect the calculation?

No, the problem specifies that the rod is "rigid and massless," so it doesn't contribute to the moment of inertia or the mass of the system.

Question

Two thin circular discs of mass 2 kg and radius 10 cm each are joined by a rigid massless rod of length 20 cm. The axis of the rod is along the perpendicular to the planes of the disc through their centres. This object is kept on a truck in such a way that the axis of the object is horizontal and perpendicular to the direction of motion of the truck.
Its friction with the floor of the truck is large enough, so that the object can roll on the truck without slipping. Take X-axis as the direction of motion of the truck and Z-axis as the vertically upwards direction. If the truck has an acceleration $9 	ext{ m/s}^2$, calculate
(a) the force of friction on each disc and
(b) the magnitude and direction of the frictional torque acting on each disc about the centre of mass $O$ of the object. Express the torque in the vector form in terms of unit vectors $\hat{i}, \hat{j}$ and $\hat{k}$ in $x, y$ and $z$-directions.

Sigma Percentile · Accepted Answer

Imagine you are standing on a truck that suddenly accelerates forward. What happens to a cylindrical object placed on the floor? Its inertia makes it want to stay behind, effectively sliding backward relative to the truck. To prevent this sliding, friction steps in, acting in the forward direction to pull the object along. This is the core physical reality we are dealing with in this problem.

### Visualizing the Setup

We have two identical discs, each of mass $m = 2 	ext{ kg}$ and radius $R = 0.1 	ext{ m}$, connected by a rigid massless rod of length $20 	ext{ cm}$. The truck accelerates in the positive X-direction with $a = 9 	ext{ m/s}^2$. The Z-axis is vertically upwards. By the right-hand rule, the Y-axis points into the page. The rod connecting the discs is aligned along this Y-axis.

### The No-Slip Condition

Let's analyze the forces on one of the discs. The only horizontal force is the forward friction $f$. This friction does two things:
1. It provides a linear acceleration $a_0$ to the center of mass: $f = m a_0$.
2. It provides a torque about the center of mass, causing an angular acceleration $\alpha$: $f R = I \alpha$.

Since the disc is a solid cylinder, its moment of inertia is $I = \frac{1}{2} m R^2$. Substituting this into the torque equation gives:

$f R = \left(\frac{1}{2} m R^2ight) \alpha \implies \alpha = \frac{2f}{mR}$

Now, here is the crucial catch: the problem states there is **no slipping**. This means the acceleration of the bottom-most point of the disc must perfectly match the truck's acceleration $a$. The bottom point has two components of acceleration: the linear acceleration $a_0$ forward, and the tangential acceleration $R \alpha$ forward due to the counter-clockwise rotation. Therefore, the no-slip condition is:

$a_{	ext{contact}} = a_0 + R \alpha = a$

### Solving for the Friction Force

Let's substitute our expressions for $a_0$ and $R \alpha$ into the no-slip condition:

$\frac{f}{m} + R \left(\frac{2f}{mR}ight) = a$

$\frac{f}{m} + \frac{2f}{m} = a \implies \frac{3f}{m} = a$

This simplifies beautifully! We can now solve directly for the friction force $f$:

$f = \frac{m a}{3} = \frac{2 	imes 9}{3} = 6 	ext{ N}$

Since it acts in the forward direction (along the X-axis), the force vector is $\mathbf{f} = 6 \hat{i} 	ext{ N}$.

### Calculating the Frictional Torque

Next, we need to find the frictional torque acting on each disc about the center of mass $O$ of the entire object. The center of mass $O$ is at the origin $(0,0,0)$.

The rod has a length of $20 	ext{ cm}$, so the discs are located at $y = -10 	ext{ cm}$ and $y = +10 	ext{ cm}$. The contact points with the truck floor are at the bottom of the discs, which is at $z = -10 	ext{ cm}$ (since the radius is $10 	ext{ cm}$). Converting to meters, the position vectors of the contact points relative to $O$ are:

For Disc 1: $\mathbf{r}_1 = -0.1 \hat{j} - 0.1 \hat{k}$
For Disc 2: $\mathbf{r}_2 = +0.1 \hat{j} - 0.1 \hat{k}$

### The Final Vector Cross Product

Torque is defined as the cross product of the position vector and the force vector: $\vec{	au} = \mathbf{r} 	imes \mathbf{f}$.

For Disc 1:
$\vec{	au}_1 = (-0.1 \hat{j} - 0.1 \hat{k}) 	imes (6 \hat{i})$
$\vec{	au}_1 = -0.6 (\hat{j} 	imes \hat{i}) - 0.6 (\hat{k} 	imes \hat{i})$

Recall your cross product rules: $\hat{j} 	imes \hat{i} = -\hat{k}$ and $\hat{k} 	imes \hat{i} = \hat{j}$. Substituting these in:

$\vec{	au}_1 = -0.6 (-\hat{k}) - 0.6 (\hat{j}) = 0.6 (\hat{k} - \hat{j}) 	ext{ N-m}$

The magnitude is $|\vec{	au}_1| = \sqrt{(-0.6)^2 + (0.6)^2} = \sqrt{0.72} \approx 0.85 	ext{ N-m}$.

For Disc 2, the process is identical, just with a positive $\hat{j}$ component in the position vector:
$\vec{	au}_2 = (0.1 \hat{j} - 0.1 \hat{k}) 	imes (6 \hat{i}) = 0.6 (-\hat{k}) - 0.6 (\hat{j}) = 0.6 (-\hat{j} - \hat{k}) 	ext{ N-m}$

The magnitude remains the same at approximately $0.85 	ext{ N-m}$. And there you have it—a perfect blend of linear dynamics, rotational kinematics, and 3D vector math!

Visualized Solution

\text{Visualizing the Setup}

\text{Equations of Motion}

\text{Angular Acceleration}

\text{No-Slip Condition}

\text{Solving for Friction}

\text{Calculating Friction Force}

\text{Position Vectors for Torque}

\text{Torque on Disc 1}

\text{Evaluating Cross Products}

\text{Magnitude of Torque 1}

\text{Torque on Disc 2}

The Sigma Insight: Rolling Motion

Visualizing the Setup

The No-Slip Condition

Solving for the Friction Force

Calculating the Frictional Torque

The Final Vector Cross Product

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