Animated Solution for Physics - Rotational Motion: A mass $m$ supported by a massless string wound around a uniform hollow cylinder of mass $m$ and radius $R$. If the string does not slip on the cylinder, with what acceleraton will the mass fall on release?

Translation Meets Rotation

Imagine you are holding a hollow cylinder mounted on an axle, with a string wrapped around it. Attached to the end of this string is a block. When you let go, the block doesn't just fall freely; it has to unspool the string, which forces the heavy cylinder to rotate. This beautiful interplay between the linear motion of the block and the rotational motion of the cylinder is the heart of this problem.

To find the acceleration of the block, we must analyze the forces and torques acting on both parts of the system simultaneously.

The Linear Perspective

Let's start by looking at the hanging block of mass $m$. There is a tug-of-war happening here. Gravity is pulling it downwards with a force of $mg$, while the tension $T$ in the string is pulling it back up. Because the block accelerates downwards with an acceleration $a$, gravity must be winning.

According to Newton's Second Law for linear motion, the net force equals mass times acceleration:

This is our first crucial equation, but it has two unknowns: $T$ and $a$. We need more information to solve it.

Torque and Inertia

Now, let's shift our focus to the hollow cylinder. The tension $T$ from the string is pulling tangentially on its edge, at a distance $R$ from the central axis. This creates a torque $\tau$ that causes the cylinder to undergo angular acceleration $\alpha$.

The torque is simply the force multiplied by the perpendicular distance:

Newton's Second Law for rotation states that the net torque equals the moment of inertia $I$ multiplied by the angular acceleration $\alpha$:

For a uniform hollow cylinder (which is geometrically equivalent to a ring), all of its mass $m$ is concentrated at the distance $R$ from the axis. Therefore, its moment of inertia is:

Substituting this into our torque equation gives:

The No-Slip Constraint

We now have two equations, but we introduced a third unknown, $\alpha$. How do we connect the linear world of the block to the rotational world of the cylinder?

The secret lies in the phrase "the string does not slip." This means that the linear acceleration $a$ of the unspooling string is exactly equal to the tangential acceleration of the cylinder's edge. Mathematically, this constraint is expressed as:

Let's substitute this constraint back into our rotational equation:

Notice how beautifully the math simplifies. One $R$ cancels out on the right side, leaving $T\cdot R=mRa$. Dividing both sides by $R$, we find a remarkably simple expression for the tension:

We have discovered that the tension required to accelerate the hollow cylinder is exactly $ma$. Now, we substitute this back into our very first linear equation:

Moving the $ma$ term to the right side, we get:

The mass $m$ cancels out from both sides, proving that the acceleration is independent of the actual mass of the objects (as long as they are equal). Dividing by 2 yields our final answer:

The block falls with exactly half the acceleration of free fall. The other half of the gravitational "effort" is spent spinning up the hollow cylinder!