Force and direction of velocity

Question

If the angle between force acting on a particle and velocity of particle is zero degree, the value of velocity will increase. If this angle is 180 degree, the value of velocity will decrease. If this angle is 90 degree, value remains constant, but direction will change. The process of changing value of velocity can be determined easily. When two same vectors act at different angle on a particle, the value and direction of resultant vector can be determined easily by using formula. How and by which formula the direction of velocity will be determined, if force acts with velocity at right angle?

Answer 1

Considering motion in a 2D plane:

If the resultant force acts perpendicular to the velocity, the force's magnitude will be equal to:

$$F = \frac{mv^2}{R}$$

Where $R$ is the radius of curvature of particle's trajectory, and $v$ is the speed.

Note that, as long as the force is perpendicular to the velocity, the speed will not increase. This is because no work is ever being done on the particle by the force. This arises from the definition of work: the product of the displacement and the component of force in the direction of the displacement. This can be expressed as:

$$W = \int^t_0 \vec{F} \cdot d\vec{r}$$

Here, the force is perpendicular to any displacement that is made, so no component of the force ever lies in the direction of the displacement, so no work is done. In the equation, the force, $\vec F$, is perpendicular to an infinitesimal displacement, $d \vec r$, at that time, so $\vec F \cdot d \vec r = 0$. Therefore $W$ is zero.

By the work-energy principle, no work means no change in kinetic energy. Therefore, the speed is constant.

The rate of change of rotation, the angular velocity, $\omega$ is defined as:

$$\omega = \frac{v}{R}$$

By substitution, it turns out that:

$$\omega = \frac{F}{mv}$$

There, we can find the angle $\theta$ that the velocity occurs in (the direction) using:

$$\theta = \theta_0+\int^t_0 \omega dt = \theta_0+\frac{1}{mv}\int^t_0 F dt$$

Where $\theta_0$ is the initial direction of velocity. For a constant magnitude for force,

$$\theta = \theta_0 + \frac{F}{mv} t$$

Note that if the magnitude of force differs, you will not get a circular orbit.

Further, this problem becomes much more complex whenever the motion not long occurs within just a 2D plane.

Answer 2

As the force acting in certain direction produce acceleration only in that direction, if force act perpendicular to the velocity it can produce an additional velocity component in perpendicular direction only. you should also consider the fact that as no force act in direction of original velocity component its value remains unchanged. Thus the resulting velocity is vector sum of these two velocities. However saying the force is acting perpendicular to velocity has many interpretations. One case is that force remains always perpendicular to velocity .This will be case of circular motion. Let me explain this: Initially you have a force acting perpendicular to velocity giving rise to a perpendicular velocity component and thus the resultant of two is a velocity turned inwards to initial velocity but at the next instant the force changes its direction so it again becomes perpendicular to this resultant velocity. Another case is the velocity is perpendicular to the initial direction of velocity and with time the perpendicular component of velocity grows and resultant of this growing perpendicular velocity and constant initial velocity component turns more and more inwards. These are the most common conditions I think usually arise and these require very basic kinematics and dynamics .

Answer 3

Pick two axes, $x$ in the direction of the velocity, $y$ perpendicular to velocity, but take care that the force vector be contained in the $x, y$ plane. Then the force makes with the velocity an angle $\theta$. The projection of the force on the velocity is $F_x = F cos(\theta)$, and the projection orthogonal to the velocity is $F_y = Fsin(\theta)$, s.t. the velocity projections are

(1) $v_x = v_0 + \int_{t_0}^t \frac {F(t) cos(\theta)}{m} dt$,

(2) $v_y = \int_{t_0}^t \frac {F(t) sin(\theta)}{m} dt$.

where I take the time at which the force was applied as $t_0$ and $v_0 = v(t_0)$ is the velocity before the force is applied. The angle $\alpha$ by which the direction of the velocity changes is given in any case by

(3) $tan (\alpha) = v_y/v_x$.

Now, if the case is that the force adjusts itself all the time so as to be perpendicular to $\vec v$, (see Phonix' question) then at the time $t = t_0 +dt$, we have

(4) $v_x(t_0 + dt) = v_0$,

(5) $v_y = \frac {F(t_0)}{m} dt$,

and $tan (\alpha)$ is according to (1)

(6) $tan (\alpha(t_0 + dt)) = \frac {F(t_0)}{m} \frac {dt}{v_0}$.

however, the force is along the direction

$\beta(t_0 + dt) = arctan(\frac {F}{m} \frac {dt}{v_0}) + \pi/2$,

or,

(7) $cotan(\beta(t_0 + dt)) = \frac {F}{m} \frac {dt}{v_0}$,

Now, we can apply the equations (1) and (2).

(8) $v_x = v_0 + \frac {F(t_0 +dt) cos(beta(t_0 + dt))}{m} dt$,

(9) $v_y = \frac {F(t_0)}{m} dt + \frac {F(t_0 + dt) sin(\theta)}{m} dt$.,

This is the tableau if the force varies I time. Maybe it can be simplified, but what I see is that the solution is iterative, because the force depends on both $t$ and $v$.

However, if the force is constant in magnitude, only always perpendicular to velocity the answer to your question is covered by the site Rotation systems. Problem interpreting an equation would be of further help.