Classical Mechanics

31.6 Non-Slip Condition

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Constraints for Rotational Motion about a Fixed Axis

A common type of problem in rotational dynamics involves objects which rotational motion is constrained by the linear motion of other objects. A typical example is when different objects are connected by ropes or ropes passing through pulleys.

Below we discuss the constraint imposed by a rope wrapped around a massive pulley of radius $R$ and connected to a hanging block. The resulting rotation of the pulley is related to the translation of the block because we will assume:

1. Non-slip condition: the rope does not slip relative to the pulley.
2. Ideal Rope: massless rope with constant length.

Consider the three points A, B and C in the rope. At time $t$, left figure, point A is in contact with the pulley and at an angle $\theta$ with respect to the horizontal, point B is the point where the rope detaches from the pulley, and point C is the point in the rope in contact with the block. The length of the portion of the rope between points A and B is the arlclength $s$, where $s=R\theta$.

If the rope does not slip relative to the pulley, point A must have the same velocity as a point at the rim of the pulley. At time $t+\mathrm{\Delta }t$ , left figure, the pulley has rotated an angle $\theta$ and point A has moved an arc of length $s$ at the same rate as a point at the rim. The velocity of point A is has a component tangent to the circle given by ${\displaystyle v_A=\frac{ds}{dt}=R\frac{d\theta }{dt}}$.

At the same time, because the rope has a constant length (ideal rope), when the pulley has rotated an angle $\theta$ the string has unwrapped a length $s$ (right figure). Consequently, points B and C and all the points in the hanging rope have moved down a length $s$. As a result, the velocity of any point in the hanging rope is the same as the one of the block equal to ${\displaystyle \frac{ds}{dt}}$.

In this problem, assumptions 1 and 2, imply that the component of the velocity of the block is:

${\displaystyle v=\frac{ds}{dt}=R\frac{d\theta }{dt}}$

After taking the derivative with respect to time, the acceleration of the block is equal the tangential component of the acceleration of a point at the rim of the pulley:

${\displaystyle a=R\alpha }$

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