The study of physics begins with a consideration of mechanical motion. In the general case, bodies move along curved trajectories with variable speeds. To describe them use the concept of acceleration. In this article, we consider what tangential and normal acceleration are.
Kinematic quantities. Speed and acceleration in physics
The kinematics of mechanical motion is a branch of physics that studies and describes the movement of bodies in space. Kinematics operates with three main quantities:
- distance traveled;
- speed;
- acceleration.
In the case of movement in a circle, similar kinematic characteristics are used, which are reduced to the central corner of the circle.
Everyone is familiar with the concept of speed. It shows the speed of change of coordinates of bodies in motion. Speed is always directed tangentially to the line along which the body moves (trajectories). Below, we denote the linear velocity by v¯ and the angular velocity by ω¯.
Acceleration is the rate of change of v¯ and ω¯. Acceleration is also a vector quantity, but its direction is completely independent of the velocity vector. Acceleration is always directed towards the force acting on the body, which causes a change in the velocity vector. The acceleration for any type of movement can be calculated by the formula:
a¯ = dv¯ / dt
The stronger the speed changes over the time interval dt, the greater the acceleration.
To understand the information set forth below, it is necessary to remember that acceleration appears as a result of any change in speed, including changes in both its module and its direction.
Tangent and Normal Acceleration
Suppose that a material point moves along some curved line. It is known that at some point in time t its velocity was equal to v¯. Since speed is a vector tangent to the trajectory, it can be represented as follows:
v¯ = v × u t ¯
Here v is the length of the vector v¯, and u t ¯ is the unit velocity vector.
To calculate the vector of full acceleration at time t, it is necessary to find the time derivative of speed. We have:
a¯ = dv¯ / dt = d (v × u t ¯) / dt
Since the velocity modulus and the unit vector change with time, using the rule of finding the derivative of the product of functions, we obtain:
a¯ = dv / dt × u t ¯ + d (u t ¯) / dt × v
The first term in the formula is called the tangential or tangent component of the acceleration, the second term is the normal acceleration.
Tangent acceleration
Once again, we write the formula for calculating the tangent acceleration:
a t ¯ = dv / dt × u t ¯
This equality means that the tangential (tangential) acceleration is directed in the same way as the velocity vector at any point on the trajectory. It numerically determines the change in velocity modulus. For example, in the case of rectilinear motion, full acceleration consists only of the tangent component. Normal acceleration with this type of displacement is zero.
The reason for the appearance of the quantity a t ¯ is the influence of an external force on a moving body.
In the case of rotation with constant angular acceleration α, the tangential component of the acceleration can be calculated by the following formula:
a t = α × r
Here r is the rotation radius of the considered material point, for which the value of a t is calculated.
Normal or centripetal acceleration
Now we write again the second component of the full acceleration:
a c ¯ = d (u t ¯) / dt × v
From geometric considerations, it can be shown that the time derivative of the unit tangent to the trajectory of the vector is equal to the ratio of the velocity modulus v to the radius r at time t. Then the expression above will be written like this:
a c = v 2 / r
This formula of normal acceleration indicates that, unlike the tangent component, it does not depend on the change in velocity, but is determined by the square of the absolute value of the velocity itself. Also, a c increases with decreasing radius of rotation at a constant value of v.
Normal acceleration is called centripetal because it is directed from the center of mass of a rotating body to the axis of rotation.
The cause of this acceleration is the central component of the force acting on the body. For example, in the case of the rotation of the planets around our sun, the centripetal force is gravitational attraction.
Normal body acceleration only changes the direction of speed. It is not able to change its module. This fact is its important difference from the tangent component of full acceleration.
Since centripetal acceleration always occurs when the velocity vector rotates, it also exists in the case of uniform rotation around a circle at which the tangential acceleration is zero.
In practice, you can feel the influence of normal acceleration if you are in the car when it makes a protracted turn. In this case, the passengers are pressed against the opposite direction of rotation of the car door. This phenomenon is the result of the action of two forces: centrifugal (displacement of passengers from their seats) and centripetal (pressure on passengers from the side of the car door).
Module and direction of full acceleration
So, we found out that the tangential component of the considered physical quantity is directed along the tangent to the trajectory of motion. In turn, the normal component is perpendicular to the trajectory at a given point. This means that the two acceleration components are perpendicular to each other. Their vector addition gives the vector of full acceleration. You can calculate its module by the following formula:
a = √ (a t 2 + a c 2 )
The direction of the vector a¯ can be determined both with respect to the vector a t ¯ and with respect to a c ¯. To do this, use the appropriate trigonometric function. For example, the angle between full and normal accelerations is:
φ = arccos (a c / a)
The solution to the problem of determining centripetal acceleration
The wheel, which has a radius of 20 cm, spins with an angular acceleration of 5 rad / s 2 for 10 seconds. It is necessary to determine the normal acceleration of the points located on the periphery of the wheel, after the specified time.
To solve the problem, we use the relationship formula between tangential and angular accelerations. We get:
a t = α × r
Since the uniformly accelerated movement lasted for a time t = 10 seconds, the linear velocity acquired during this time was equal to:
v = a t × t = α × r × t
We substitute the resulting formula in the corresponding expression for normal acceleration:
a c = v 2 / r = α 2 × t 2 × r
It remains to substitute the known values into this equality and write down the answer: a c = 500 m / s 2 .