Animated Solution for Physics - Rotational Motion: A mass $m$ hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass $m$ and radius $R$. Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass $m$, if the string does not slip on the pulley, is

This classic problem beautifully bridges the gap between translational kinematics and rotational dynamics. It challenges us to look beyond the idealized "massless pulley" and consider the real-world implications of a pulley having its own mass and rotational inertia.

Imagine the system in motion. We have a block of mass $m$ hanging from a string. This string is wrapped around a pulley, which is a uniform circular disc also of mass $m$ and radius $R$. When the block is released, it accelerates downwards. However, it doesn't fall as fast as it would in free fall. Why? Because the gravitational force acting on the block must not only accelerate the block linearly but also accelerate the pulley rotationally.

To solve this, we must break the system down into two distinct parts: the translational motion of the hanging block and the rotational motion of the pulley.

Let's isolate the hanging block. There are two forces acting on it: 1. The downward force of gravity, $mg$. 2. The upward tension force from the string, $T$.

Since the block is accelerating downwards with an acceleration $a$, we apply Newton's Second Law ($F_{\text{net}}=ma$) in the downward direction:

Now, let's shift our focus to the pulley. The string pulls tangentially on the edge of the pulley with tension $T$. This force creates a torque ($\tau$) that causes the pulley to rotate. The torque is the product of the force and the perpendicular distance from the axis of rotation, which is the radius $R$.

According to Newton's Second Law for rotation, torque equals the moment of inertia ($I$) multiplied by the angular acceleration ($\alpha$):

To solve our equations, we need to link the translational variables with the rotational variables. We are given two vital pieces of information:

1. The Pulley is a Uniform Disc: The moment of inertia of a uniform solid disc rotating about its central axis is a standard result:

2. The String Does Not Slip: This "no-slip" condition is the physical bridge between the two motions. It dictates that the linear acceleration of the string ($a$) is exactly equal to the tangential acceleration of the rim of the pulley. Mathematically, this is expressed as:

Let's substitute our constraints into the rotational equation $(2)$:

Notice the elegance of the physics here: the radius $R$ cancels out completely! This tells us that the tension required to accelerate the pulley depends on its mass and linear acceleration, but not its size.

Now, we take this expression for tension and substitute it back into our translational equation $(1)$:

We group the acceleration terms on one side:

Finally, the mass $m$ cancels out from both sides, leaving us with the final acceleration:

The result $a=\frac{2}{3}g$ is deeply satisfying. If the pulley were massless, the acceleration would simply be $g$. However, because the pulley has mass and rotational inertia, it "steals" some of the driving force to spin itself up. The factor of $\frac{2}{3}$ perfectly quantifies this resistance to rotational motion for a uniform disc.