The body gets accelerated if. Normal acceleration
When bodies move, their velocities usually change either in absolute value, or in direction, or simultaneously both in absolute value and in direction.
If you throw a stone at an angle to the horizon, then its speed will change both in magnitude and in direction.
The change in the speed of the body can occur both very quickly (movement of a bullet in the bore when fired from a rifle), and relatively slowly (movement of a train when it is sent). To be able to find the speed at any moment of time, it is necessary to enter a value that characterizes the rate of change of speed. This value is calledacceleration.
- this is the ratio of the change in speed to the period of time during which this change occurred. The average acceleration can be determined by the formula:
where - acceleration vector .
The direction of the acceleration vector coincides with the direction of the change in speed Δ = - 0 (here 0 is the initial speed, that is, the speed at which the body began to accelerate).
At time t1 (see Figure 1.8) the body has a speed of 0 . At time t2 the body has a speed. According to the vector subtraction rule, we find the vector of speed change Δ = - 0 . Then the acceleration can be defined as follows:
Rice. 1.8. Average acceleration.
in SI unit of acceleration is 1 meter per second per second (or meter per second squared), that is
A meter per second squared is equal to the acceleration of a point moving in a straight line, at which in one second the speed of this point increases by 1 m / s. In other words, acceleration determines how much the speed of a body changes in one second. For example, if the acceleration is 5 m / s 2, then this means that the speed of the body increases by 5 m / s every second.
Definition
body acceleration called a vector quantity showing the rate of change in the speed of a body. Designate the acceleration as $\overline(a)$.
Average body acceleration
Let us assume that at times $t$ and $t+\Delta t$ the velocities are equal to $\overline(v)(t)$ and $\overline(v)(t+\Delta t)$. It turns out that during the time $\Delta t$ the speed changes by:
\[\Delta \overline(v)=\overline(v)\left(t+\Delta t\right)-\overline(v)\left(t\right)\left(1\right),\]
then the average acceleration of the body is:
\[\left\langle \overline(a)\right\rangle \left(t,\ t+\Delta t\right)=\frac(\Delta \overline(v))(\Delta t)\left(2\ right).\]
instant body acceleration
Let's set the time interval $\Delta t$ to zero, then from equation (2) we get:
\[\overline(a)=(\mathop(\lim )_(\Delta t\to 0) \frac(\Delta \overline(v))(\Delta t)=\frac(d\overline(v) )(dt)\left(3\right).\ )\]
Formula (3) is the definition of instantaneous acceleration. Whereas, in a Cartesian coordinate system:
\[\overline(r)=x\left(t\right)\overline(i)+y\left(t\right)\overline(j)+z\left(t\right)\overline(k)\ left(4\right),\ a\ \overline(v)=\frac(d\overline(r))(dt)(5)\]
we get:
\[\overline(a)=\overline(i)\frac(d^2x)(dt^2)+\overline(j)\frac(d^2y)(dt^2)+\overline(k)\ frac(d^2z)(dt^2)=\frac(d^2\overline(r))(dt^2)\left(6\right).\]
From expression (6) it follows that the acceleration projections on the coordinate axes (X,Y,Z) are equal to:
\[\left\( \begin(array)(c) a_x=\frac(d^2x)(dt^2), \\ a_y=\frac(d^2y)(dt^2) \\ a_z=\ frac(d^2z)(dt^2).\end(array)\right.(7),\]
In this case, we find the acceleration module in accordance with the expression:
To clarify the question of the direction of acceleration of the body's motion, we represent the velocity vector as:
\[\overline(v)=v\overline(\tau )\left(8\right),\]
where $v$ is the modulus of the body's velocity; $\overline(\tau )$ - unit vector tangent to the trajectory of the material point. We substitute expression (8) into the definition of instantaneous speed, we get:
\[\overline(a)=(\frac(d\overline(v))(dt) =\frac(d)(dt)\left(v\overline(\tau )\right)=\overline(\tau )\frac(dv)(dt)+v\frac(d\overline(\tau ))(dt)\left(9\right).\ )\]
The unit tangent vector $\overline(\tau )$ is defined by a trajectory point, which in turn is characterized by a distance ($s$) from the starting point. So the vector $\overline(\tau )$ is a function of $s$:
\[\overline(\tau )=\overline(\tau )\left(s\right)\left(10\right).\]
Parameter $s$ is a function of time. We get:
\[\frac(d\overline(\tau ))(dt)=\frac(d\overline(\tau ))(ds)\frac(ds)(dt)\left(11\right),\]
where the vector $\overline(\tau )$ does not change modulo. This means that the vector $\frac(d\overline(\tau ))(ds)$ is perpendicular to $\overline(\tau )$. The vector $\overline(\tau )(\rm \ )$ is tangent to the trajectory, $\frac(d\overline(\tau ))(ds)$ is perpendicular to this tangent, that is, it is directed along the normal, which is called the main . The unit vector in the direction of the main normal will be denoted by $\overline(n)$.
The value $\left|\frac(d\overline(\tau ))(ds)\right|=\frac(1)(R)$, where $R$ is the radius of curvature of the trajectory.
And so we got:
\[\frac(d\overline(\tau ))(ds)=\frac(\overline(n))(R)\left(12\right).\]
Taking into account that $\frac(ds)(dt)=v$, from (9) we can write the following:
\[\overline(a)=\overline(\tau )\frac(dv)(dt)+v\frac(\overline(n))(R)v=\overline(\tau )\frac(dv)( dt)+\frac(v^2)(R)\overline(n)\left(13\right).\]
Expression (13) shows that the total acceleration of the body consists of two components that are mutually perpendicular. Tangential acceleration ($(\overline(a))_(\tau )$) directed tangentially to the motion trajectory and equal to:
\[(\overline(a))_(\tau )=\overline(\tau )\frac(dv)(dt)(14)\]
and normal (centripetal) acceleration ($(\overline(a))_n$) directed perpendicular to the tangent to the trajectory at the point where the body is located along the main normal (to the center of curvature of the trajectory) and equal to:
\[(\overline(a))_n=\frac(v^2)(R)\overline(n)\left(15\right).\]
The total acceleration modulus is:
The unit of acceleration in the International System of Units (SI) is meter per second squared:
\[\left=\frac(m)(s^2).\]
Rectilinear body movement
If the trajectory of the material point is a straight line, then the acceleration vector is directed along the same straight line as the velocity vector. Only the speed is changed.
Variable motion is called accelerated if the speed of a material point is constantly increasing in absolute value. In this case, $a>0$, the vectors of acceleration and velocity are co-directed.
If the modulo speed decreases, then the movement is called slow ($a
The motion of a material point is called equally variable and rectilinear if the motion occurs with constant acceleration ($\overline(a)=const$). With uniformly variable motion, the instantaneous speed ($\overline(v)$) and the acceleration of a material point are related by the expression:
\[\overline(v)=(\overline(v))_0+\overline(a)t\ \left(3\right),\]
where $(\overline(v))_0$ is the speed of the body at the initial moment of time.
Examples of problems with a solution
Example 1
Exercise: The motions of two material points are given by the following kinematic equations: $x_1=A+Bt-Ct^2$ and $x_2=D+Et+Ft^2,$ which are the accelerations of these two points at the time when their velocities are equal, if $ A$, B,C,D,E.F - constants greater than zero.
Solution: Find the acceleration of the first material point:
\[(a_1=a)_(x1)=\frac(d^2x_1)(dt^2)=\frac(d^2)(dt^2)\left(A+Bt-Ct^2\right) =-2C\ (\frac(m)(c^2)).\]
At the second material point, the acceleration will be equal to:
\[(a_2=a)_(x2)=\frac(d^2x_2)(dt^2)=\frac(d^2)(dt^2)\left(D+Et+Ft^2\right) =2F\left(\frac(m)(c^2)\right).\]
We got that the points move with constant accelerations, which do not depend on time, so it is not necessary to look for the moment in time at which the speeds are equal.
Answer:$a_1=-2C\frac(m)(c^2)$, $a_2=2F\frac(m)(c^2)$
Example 2
Exercise: The motion of a material point is given by the equation: $\overline(r)\left(t\right)=A\left(\overline(i)(\cos \left(\omega t\right)+\overline(j)(\sin \left(\omega t\right)\ )\ )\right),$ where $A$ and $\omega $ are constants. Draw the trajectory of the point, depict on it the acceleration vector of this point. What is the modulus of centripetal acceleration ($a_n$) of the point in this case?
Solution: Consider the equation of motion of our point:
\[\overline(r)\left(t\right)=A\left(\overline(i)(\cos \left(\omega t\right)+\overline(j)(\sin \left(\omega t\right)\ )\ )\right)\ \left(2.1\right).\]
In the coordinate notation, equation (2.1) corresponds to the system of equations:
\[\left\( \begin(array)(c) x\left(t\right)=A(\rm cos)\left(\omega t\right), \\ y(t)=A(\sin \left(\omega t\right)\ ) \end(array) \left(2.2\right).\right.\]
We square each equation of system (2.2) and add them:
We have obtained the equation for a circle of radius $A$ (Fig.1).
The value of centripetal acceleration, given that the radius of the trajectory is equal to A, we find as:
Velocity projections on the coordinate axes are:
\[\left\( \begin(array)(c) v_x=\frac(dx\left(t\right))(dt)=-A\ \omega \ (\rm sin)\left(\omega t\ right), \\ v_y=\frac(dy\left(t\right))(dt)=A(\omega \ \cos \left(\omega t\right)\ ) \end(array) \left(2.5 \right).\right.\]
The speed value is:
Substitute the result (2.6) into (2.4), the normal acceleration is:
It is easy to show that the movement of a point in our case is a uniform movement along a circle and the total acceleration of the point is equal to the centripetal acceleration. To do this, you can take the derivative of the projections of velocities (2.5) with respect to time and use the expression:
get:
Answer:$a_n=A(\omega )^2$
In this lesson, we will consider an important characteristic of uneven movement - acceleration. In addition, we will consider non-uniform motion with constant acceleration. This movement is also called uniformly accelerated or uniformly slowed down. Finally, we will talk about how to graphically depict the speed of a body as a function of time in uniformly accelerated motion.
Homework
Solving problems for this lesson, you will be able to prepare for questions 1 of the GIA and questions A1, A2 of the exam.
1. Tasks 48, 50, 52, 54 sb. tasks of A.P. Rymkevich, ed. ten.
2. Write down the dependences of the speed on time and draw graphs of the dependence of the speed of the body on time for the cases shown in fig. 1, cases b) and d). Mark the turning points on the graphs, if any.
3. Consider the following questions and their answers:
Question. Is the acceleration free fall acceleration, according to the definition given above?
Answer. Of course it is. Free fall acceleration is the acceleration of a body that falls freely from a certain height (air resistance must be neglected).
Question. What happens if the acceleration of the body is directed perpendicular to the speed of the body?
Answer. The body will move uniformly in a circle.
Question. Is it possible to calculate the tangent of the angle of inclination using a protractor and a calculator?
Answer. Not! Because the acceleration obtained in this way will be dimensionless, and the dimension of acceleration, as we showed earlier, must have the dimension of m/s 2 .
Question. What can be said about motion if the graph of speed versus time is not a straight line?
Answer. We can say that the acceleration of this body changes with time. Such a movement will not be uniformly accelerated.
Translational and rotational movements
Translational this movement is called solid body, at which any straight line drawn in this body moves, remaining parallel to its initial direction.
Translational motion should not be confused with rectilinear. During the translational motion of the body, the trajectories of its points can be any curved lines.
Rotational motion of a rigid body around fixed axle is called such a movement in which any two points belonging to the body (or invariably associated with it) remain motionless during the entire movement
Speed is the ratio of the distance traveled to the time it took to travel the distance.
The speed is the same is the sum of initial velocity and acceleration multiplied by time.
Speed is the product of the angular velocity and the radius of the circle.
v=S/t
v=v 0 +a*t
v=ωR
Acceleration of a body in uniformly accelerated motion- a value equal to the ratio of the change in speed to the time interval during which this change occurred.
Tangential (tangential) acceleration is the component of the acceleration vector directed along the tangent to the trajectory at a given point in the trajectory. Tangential acceleration characterizes the change in speed modulo at curvilinear motion.
Rice. 1.10. tangential acceleration.
The direction of the tangential acceleration vector τ (see Fig. 1.10) coincides with the direction of the linear velocity or is opposite to it. That is, the tangential acceleration vector lies on the same axis as the tangent circle, which is the trajectory of the body.
Normal acceleration is a component of the acceleration vector directed along the normal to the motion trajectory at a given point on the body motion trajectory. That is, the normal acceleration vector is perpendicular to the linear speed of movement (see Fig. 1.10). Normal acceleration characterizes the change in speed in the direction and is denoted by the letter n. The normal acceleration vector is directed along the radius of curvature of the trajectory.
Full acceleration in curvilinear motion, it consists of tangential and normal accelerations along vector addition rule and is determined by the formula:
(according to the Pythagorean theorem for a rectangular rectangle).
The direction of full acceleration is also determined vector addition rule:
angular velocity is called a vector quantity equal to the first derivative of the angle of rotation of the body with respect to time:
v=ωR
angular acceleration is called a vector quantity equal to the first derivative of the angular velocity with respect to time:
Fig.3
When the body rotates around a fixed axis, the angular acceleration vector ε is directed along the axis of rotation towards the vector of the elementary increment of the angular velocity. With accelerated movement, the vector ε co-directed to the vector ω (Fig. 3), when slowed down, it is opposite to it (Fig. 4).
Fig.4
Tangential acceleration component a τ =dv/dt , v = ωR and
Normal component of acceleration
This means that the relationship between linear (path length s, traveled by a point along an arc of radius R, linear velocity v, tangential acceleration a τ, normal acceleration a n) and angular quantities (angle of rotation φ, angular velocity ω, angular acceleration ε) is expressed as follows formulas:
s = Rφ, v = Rω, and τ = R?, a n = ω 2 R.
In the case of equally variable motion of a point along a circle (ω=const)
ω = ω 0 ± ?t, φ = ω 0 t ± ?t 2 /2,
where ω 0 is the initial angular velocity.
And why is it needed. We already know what a frame of reference is, the relativity of motion and material point. Well, it's time to move on! Here we will review the basic concepts of kinematics, bring together the most useful formulas on the basics of kinematics, and give a practical example of solving the problem.
Let's solve the following problem: A point moves in a circle with a radius of 4 meters. The law of its motion is expressed by the equation S=A+Bt^2. A=8m, B=-2m/s^2. At what point in time is the normal acceleration of a point equal to 9 m/s^2? Find the speed, tangential and total acceleration of the point for this moment in time.
Solution: we know that in order to find the speed, we need to take the first time derivative of the law of motion, and the normal acceleration is equal to the private square of the speed and the radius of the circle along which the point moves. Armed with this knowledge, we find the desired values.
Need help solving problems? A professional student service is ready to provide it.