Motion is one of the important properties of matter in our universe. Indeed, even at an absolute zero temperature, the movement of particles of matter does not stop completely. In physics, motion is described by a number of parameters, the main of which is acceleration. In this article we will reveal in more detail the question regarding what tangential acceleration is and how to calculate it.
Acceleration in Physics
Acceleration is understood to mean the speed with which the speed of the body changes during its movement. Mathematically, this definition is written as follows:
a¯ = d v¯ / dt
This is the kinematic definition of acceleration. The formula shows that it is calculated in meters per square second (m / s 2 ). Acceleration is a vector feature. Its direction has nothing to do with the direction of speed. Acceleration is directed towards the change of speed. Obviously, in the case of uniform movement in a straight line, there is no change in speed, so the acceleration is zero.
If we talk about acceleration as a magnitude of dynamics, then we should recall Newton's law:
F¯ = m × a¯ =>
a¯ = F¯ / m
The cause of the a¯ value is the force F¯ acting on the body. Since the mass m is a scalar quantity, the acceleration is directed towards the action of the force.
Trajectory and full acceleration
Speaking about acceleration, speed and the distance traveled, one should not forget about another important characteristic of any movement - the trajectory. It is understood as an imaginary line along which the studied body moves. In general, it can be curved or straight. The most common curve path is a circle.
Suppose that the body moves along a trajectory curve. Moreover, its speed varies according to some law v = v (t). At any point on the trajectory, the velocity is directed tangentially to it. Speed can be expressed as the product of its module v and the elementary vector u¯. Then for acceleration we get:
v¯ = v × u¯;
a¯ = d v¯ / dt = d (v × u¯) / dt
Applying the rule of calculating the derivative of the product of functions, we obtain:
a¯ = d (v × u¯) / dt = dv / dt × u¯ + v × d u¯ / dt
Thus, the full acceleration a¯ when moving along a trajectory curve is decomposed into two components. In this article we consider in detail only the first term, which is called the tangential acceleration of a point. As for the second term, we only say that it is called normal acceleration and is directed to the center of curvature.
Tangential acceleration
Denote this component of full acceleration by a t ¯. We rewrite the tangential acceleration formula:
a t ¯ = dv / dt × u¯
What does this equality mean? First, the component a t ¯ characterizes the change in the absolute value of the velocity, without taking its direction into account. So, in the process of motion, the velocity vector can be constant (rectilinear) or constantly change (curvilinear), but if the velocity modulus remains unchanged, then a t ¯ will be equal to zero.
Secondly, tangential acceleration is directed in exactly the same way as the velocity vector. This fact is confirmed by the presence of a factor in the above formula in the form of an elementary vector u¯. Since u¯ is directed along the tangent to the trajectory, the component a t ¯ is often called tangent acceleration.
Based on the definition of tangential acceleration, we can conclude: the quantities a¯ and a t ¯ always coincide in the case of rectilinear movement of bodies.
Tangent and angular acceleration when moving around a circle
We found out above that movement along any curved path leads to the appearance of two acceleration components. One type of movement along a curved line is the rotation of bodies and material points around a circle. This type of movement is conveniently described by angular characteristics such as angular acceleration, angular velocity and angle of rotation.
By angular acceleration α is understood the magnitude of the change in the velocity of angular ω:
α = d ω / dt
Angular acceleration leads to an increase in speed. Obviously, in this case, the linear velocity of each point that participates in the rotation increases. Therefore, there must be an expression that relates angular and tangential acceleration. We will not go into details of the output of this expression, but give it right away:
a t = α × r
The values of a t and α are directly proportional to each other. In addition, a t increases with increasing distance r from the axis of rotation to the point in question. That is why, during rotation, it is convenient to use α rather than a t (α does not depend on the radius of rotation r).
Task example
It is known that a material point rotates around an axis with a radius of 0.5 meters. Its angular velocity in this case varies according to the following law:
ω = 4 × t + t 2 + 3
It is necessary to determine with what tangential acceleration the point will rotate at 3.5 seconds.
To solve this problem, you must first use the formula for angular acceleration. We have:
α = d ω / dt = 2 × t + 4
Now we should apply the equality that relates the quantities a t and α, we obtain:
a t = α × r = t + 2
When writing the last expression, we substituted the value r = 0.5 m from the condition. As a result, we got a formula according to which tangential acceleration depends on time. Such circular motion is not uniformly accelerated. To obtain an answer to the problem, it remains to substitute a known moment in time. We get the answer: a t = 5.5 m / s 2 .