1
$\begingroup$

When a body moves in a circular loop, there are basically two types of acceleration acting on the object. One is the linear acceleration which is basically tangent to the circle, and there is an acceleration that acts inward which is basically centripetal acceleration.
These two acceleration vectors are perpendicular to each other and they have a net acceleration vector pointing inward.

My question is since there is a net acceleration due to these two acceleration vectors then the net force must also be acting in that direction. Then why does the object not move in the direction of that net force rather than moving along the circular path. Please understand that I am not talking about the centripetal force, rather i am taking about the net resultant force(centripetal + tangential force).

I feel that along the tangential direction of my circular path, the force would be basically the force of my engine which is driving my car. This force kind of propels me forward in the linear direction, and during the circular turn, the friction force basically would provide me the centripetal force and prevent me from flying off my turn.

So, what would be the meaning of the resultant of these two forces would imply in my real world because the resultant of these two force would be parallel to the X-axis and would act inward. I am basically moving the circular arc due to the combined effects of these two forces. Please I am standard 10 student, so my physics concepts are not really very advanced. Can someone explain this to me in a little conceptual manner?

$\endgroup$
  • $\begingroup$ Woah, wall of text! But if you take a point at an instant and draw the resultant force, isn't the car going in that direction the next moment? (as opposed to going in the direction the engines are pushing it and the direction the centripetal force is pulling it) $\endgroup$ – Jerry May 11 '13 at 20:52
  • $\begingroup$ If thats the case, then wont the car move straight linearly along the direction of the net force. How is it able to still move along the circular turn? $\endgroup$ – Raj May 11 '13 at 21:14
  • $\begingroup$ The direction of the net force changes as soon as the direction of the car changes, and the net direction of that net force becomes a circular path. $\endgroup$ – Jerry May 12 '13 at 7:04
  • $\begingroup$ wrong to say that 'acceleration acting on the object'. we may cause Forces to act on a body and may accelerate the body in some direction which can be broken down into x-y axes or x-y-z axes. $\endgroup$ – KawaiKx Sep 24 '14 at 9:03
1
$\begingroup$

The initial premise of your question is not, in general, correct.

Consider a 1000 kg car driving at a constant speed of 20 m/s around a flat circular racetrack with a radius of 500 meters. Note the constant speed: the car will take the same time to drive around the track, 157.1 seconds, hour after hour. The only change in the car's velocity is in the direction of the velocity, not its size. The only acceleration is the centripetal acceleration, and the only horizontal force needed is the centripetal force $$a_{cp}=\frac {v^2}{r}=0.8 \frac{m}{sec^2}$$ $$F_{cp}=m \times a_{cp}=800 \ newtons$$

The force exerted by the car's engine serves only to balance the various drag forces.

This force is supplied by the friction between the tires and the pavement; pour out some oil on the track to see what happens when the required centripetal force is not present!

The only acceleration is directed exactly towards the center of the circular track; there is no tangential acceleration. The car, at some moment, is travelling North at 20 m/s, and one half-circuit later, it is travelling South at the same speed. Clearly, it has accelerated.

Assume now that the driver presses on the brake pedal in a manner that she knows will bring the car to a stop in 40 seconds. There is now a tangential acceleration, at $-0.5 \frac{m}{sec^2}$, in addition to the centripetal acceleration above, and a tangential force of 500 newtons directed towards the back of the car. The total force needed from the tires is now the resultant of these two forces: 943.4 newtons directed around 32 degrees aft of inward. Touching the brakes could throw you into a skid! Of course, as the braking changes the car's speed, the centripetal force will decrease...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.