All the bodies that surround us are in constant motion. Moving in the space of bodies is observed at all scale levels, starting with the movement of elementary particles in the atoms of matter and ending with the accelerated movement of galaxies in the Universe. In any case, the process of movement occurs with acceleration. In this article, we consider in detail the concept of tangent acceleration and give a formula by which it can be calculated.
Kinematic quantities
Before we talk about tangential acceleration, we consider what quantities are customary to characterize the arbitrary mechanical movement of bodies in space.
First of all, this is the path L. It shows what distance, in meters, centimeters, kilometers, and so on, the body has traveled in a certain period of time.
The second important characteristic in kinematics is the speed of the body. Unlike the path, it is a vector quantity and is directed along the path of the body. Speed determines the speed at which spatial coordinates change over time. The formula for its calculation is:
v¯ = dL / dt
Speed is the time derivative of the path.
Finally, the third important characteristic of the motion of bodies is acceleration. According to the definition in physics, acceleration is a quantity that determines the change in speed with time. The formula for it can be written as:
a¯ = dv¯ / dt
Acceleration, like speed, is also a vector quantity, but unlike it, it is directed towards a change in speed. The direction of acceleration also coincides with the vector of the resulting force exerting an effect on the body.
Trajectory and acceleration
Many problems in physics are considered in the framework of rectilinear motion. In this case, as a rule, they do not talk about the tangent acceleration of a point, but work with linear acceleration. However, if the movement of the body is not linear, then its full acceleration can be decomposed into two components:
In the case of linear motion, the normal component is equal to zero; therefore, they do not talk about vector decomposition of acceleration.
Thus, the trajectory of motion largely determines the nature and components of full acceleration. Under the trajectory of movement is understood an imaginary line in space along which the body moves. Any curved path leads to the appearance of nonzero acceleration components noted above.
Determination of tangential acceleration
Tangential or, as it is also called, tangential acceleration is a component of full acceleration, which is directed along the tangent to the trajectory of motion. Since the velocity is also directed along the trajectory, the tangential acceleration vector coincides with the velocity vector.
Above was given the concept of acceleration as a measure of speed change. Since speed is a vector, it can be changed either modulo or direction. Tangent acceleration determines only a change in the speed modulus.
Note that in the case of rectilinear motion, the velocity vector does not change its direction, therefore, in accordance with the above definition, tangential acceleration and linear acceleration are one and the same quantity.
Obtaining the equation of tangent acceleration
Suppose that the body moves along a certain curve of the trajectory. Then its velocity v¯ at the selected point can be represented as follows:
v¯ = v * u t ¯
Here v is the module of the vector v¯, u t ¯ is the unit velocity vector directed along the tangent to the trajectory.
Using the mathematical definition of acceleration, we obtain:
a¯ = dv¯ / dt = d (v * u t ¯) / dt = dv / dt * u t ¯ + v * d (u t ¯) / dt
When finding the derivative, the property of the product of two functions was used here. We see that the full acceleration a¯ at the point in question corresponds to the sum of two terms. They are tangent and normal point acceleration, respectively.
Let's say a few words about normal acceleration. It is responsible for changing the velocity vector, that is, for changing the direction of motion of the body along the curve. If we explicitly calculate the value of the second term, then we get the formula for normal acceleration:
a n = v * d (u t ¯) / dt = v 2 / r
Normal acceleration is directed along the normal restored to a given point in the curve. In the case of circular motion, normal acceleration is centripetal.
The equation of tangent acceleration a t ¯ has the form:
a t ¯ = dv / dt * u t ¯
This expression suggests that the tangential acceleration does not correspond to a change in direction, but the modulus of velocity v¯ for a moment in time. Since the tangential acceleration is directed along the tangent to the considered point of the trajectory, it is always perpendicular to the normal component.
Tangential acceleration and full acceleration module
Above, all the information was presented that allows you to calculate the full acceleration through the tangent and normal. Indeed, since both components are mutually perpendicular, their vectors form the legs of a right triangle, the hypotenuse of which is the vector of full acceleration. This fact allows us to write the formula for the full acceleration module in the following form:
a = √ (a n 2 + a t 2 )
The angle θ between full acceleration and tangential can be defined as follows:
θ = arccos (a t / a)
The greater the tangential acceleration, the closer the directions of tangent and full acceleration.
The relationship of tangent and angular acceleration
A typical curvilinear path along which bodies move in technology and nature is a circle. Indeed, the movement of gears, blades and planets around its own axis or around its bodies occurs precisely around the circumference. The motion corresponding to this trajectory is called rotation.
The kinematics of rotation are characterized by the same values as the kinematics of motion in a straight line, however, they are angular in nature. So, to describe the rotation using the Central angle of rotation θ, angular velocity ω and acceleration α. The following formulas are valid for these quantities:
ω = dθ / dt;
α = dω / dt
Suppose that the body has made one revolution around the axis of rotation in time t, then for the angular velocity we can write:
ω = 2 * pi / t
Linear speed in this case will be equal to:
v = 2 * pi * r / t
Where r is the radius of the trajectory. The last two expressions allow you to write the formula for the connection of two speeds:
v = ω * r
Now we calculate the time derivative of the left and right sides of the equality, we get:
dv / dt = r * dω / dt
On the right side of the equality is the product of angular acceleration by the radius of the circle. The left side of the equality is a change in the velocity modulus, that is, tangential acceleration.
Thus, tangential acceleration and a similar angular quantity are related by the equality:
a t = α * r
If we assume that the disk rotates, then the tangential acceleration of the point at a constant value of α will increase linearly with increasing distance from this point to the axis of rotation r.
Next, we solve two problems for the application of the above formulas.
Determination of tangential acceleration from a known velocity function
It is known that the speed of a body that moves along a certain curve of the trajectory is described by the following function of time:
v = 2 * t 2 + 3 * t + 5
It is necessary to determine the formula of tangent acceleration and find its value at time t = 5 seconds.
First, we write the formula for the tangential acceleration module:
a t = dv / dt
That is, to calculate the function a t (t), one should determine the time derivative of the velocity. We have:
a t = d (2 * t 2 + 3 * t + 5) / dt = 4 * t + 3
Substituting the time t = 5 seconds into the obtained expression, we arrive at the answer: a t = 23 m / s 2 .
Note that the graph of speed versus time in this problem is parabola, while the graph of tangential acceleration is a straight line.
The problem of determining tangential acceleration
It is known that the material point of the beginning of uniformly accelerated rotation from the zero moment of time. 10 seconds after the start of rotation, its centripetal acceleration became equal to 20 m / s 2 . It is necessary to determine the tangent acceleration of the point after 10 seconds, if it is known that the radius of rotation is 1 meter.
First, we write the formula for centripetal or normal acceleration a c :
a c = v 2 / r
Using the relationship between linear and angular velocity, we obtain:
a c = ω 2 * r
With uniformly accelerated motion, the velocity with angular acceleration are related by the formula:
ω = α * t
Substituting ω into the equality for a c , we obtain:
a c = α 2 * t 2 * r
Linear acceleration through tangential is expressed as follows:
α = a t / r
Substituting the last equality in the penultimate, we get:
a c = a t 2 / r 2 * t 2 * r = a t 2 / r * t 2 =>
a t = √ (a c * r) / t
The last formula, taking into account the data from the conditions of the problem, leads to the answer: a t = 0.447 m / s 2 .