Consider the motion of a point on a circular trajectory. The acceleration in a linear motion (a) and the acceleration in angular motion (α), are related as : (Take r as the radius of circular trajectory)

Question

Prepp Team · Accepted Answer

Circular Motion Acceleration

In physics, when a point moves along a circular path, we can describe its motion using both linear quantities and angular quantities. Linear quantities include linear displacement, linear velocity, and linear acceleration. Angular quantities include angular displacement, angular velocity, and angular acceleration.

The question asks about the relationship between linear acceleration (denoted as $a$) and angular acceleration (denoted as $\alpha$) for a point on a circular trajectory with radius $r$.

First, let's recall the relationship between linear velocity ($v$) and angular velocity ($\omega$) for a point moving in a circle of radius $r$:

$$ v = r\omega $$

Linear acceleration is the rate of change of linear velocity with respect to time. Angular acceleration is the rate of change of angular velocity with respect to time.

To find the relationship between linear acceleration and angular acceleration, we can differentiate the linear velocity equation with respect to time ($t$).

$$ \frac{dv}{dt} = \frac{d}{dt}(r\omega) $$

Since the radius $r$ of the circular trajectory is constant, we can take $r$ out of the differentiation:

$$ \frac{dv}{dt} = r \frac{d\omega}{dt} $$

By definition, the rate of change of linear velocity is linear acceleration ($a$), i.e., $a = \frac{dv}{dt}$. The rate of change of angular velocity is angular acceleration ($\alpha$), i.e., $\alpha = \frac{d\omega}{dt}$.

Substituting these definitions into the equation, we get:

$$ a = r\alpha $$

This relationship specifically refers to the tangential component of linear acceleration. When a point is accelerating or decelerating along a circular path, its linear speed changes. The rate of change of linear speed is the tangential acceleration, which is directly related to the angular acceleration by $a_t = r\alpha$. The question uses $a$ generally for linear acceleration in the context of relation with $\alpha$, implying this tangential component.

Therefore, the relationship between linear acceleration ($a$) and angular acceleration ($\alpha$) for a point on a circular trajectory of radius $r$ is $a = r\alpha$.

Prepp Team · Answer

a = r2 &alpha;

Prepp Team · Answer

a = r &alpha;2

Prepp Team · Answer

a = $r\over\alpha$

Circular Motion Acceleration

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