Circular Motion Force at Point A: Stunt Car on Frictionless Track

[Help1212]

Homework Statement

In the figure, a stunt car driver negotiates the frictionless track shown in such a way that the car is barely on the track at the top of the loop. The radius of the track is 9.9 m and the mass of the car is 1800 kg. Find the force of the car on the track when the car is at point A

http://session.masteringphysics.com/problemAsset/1107086/1/7.10.jpg"

Homework Equations

Accelaration(radius) V²/R F=ma

The Attempt at a Solution

massxacceleration =1800kg x V²/9.9m ... v²=1800kg/9.9 =181 .. v=13.4 m/s

F=1800kg x (13.4)²/9.9

Can someone help me please.. what am I doing wrong?

 

[WatermelonPig]

Is the loop vertical? Is the car traveling on an inclined slope?

 

[Help1212]

sorry for the confusion i included a picture link for the problm

 

[gneill]

If the track is frictionless then his trajectory is ballistic; he can't use the engine to accelerate because the tires have no friction with the track.

The condition posed is that the car is "barely on the track at the top of the loop". So work out its required speed there. Then use whatever conservation laws spring to mind.

 

[rwisz]

I would like to clarify that circular motion is a complex concept that involves not only the mass and acceleration of an object, but also the centripetal force acting on the object. In this scenario, the car is moving in a circular path, which means it is constantly changing direction and therefore experiencing a centripetal acceleration towards the center of the loop. This acceleration is caused by a force known as the centripetal force, which is provided by the track in this case.

To find the force of the car on the track at point A, we need to use the equation F=ma, where m is the mass of the car and a is the centripetal acceleration. The centripetal acceleration can be calculated using the equation a=v^2/r, where v is the speed of the car at point A and r is the radius of the loop.

Using the given information, we can calculate the speed of the car at point A using the equation v=sqrt(gr), where g is the acceleration due to gravity and r is the radius of the loop. Plugging in the values, we get v=sqrt(9.8 m/s^2 x 9.9 m)= 9.9 m/s.

Now, we can calculate the centripetal acceleration using the equation a=v^2/r, which gives us a=(9.9 m/s)^2 / 9.9 m= 9.9 m/s^2.

Finally, we can calculate the force of the car on the track at point A using the equation F=ma, which gives us F=1800 kg x 9.9 m/s^2= 17,820 N.

Therefore, the force of the car on the track at point A is 17,820 N. It is important to note that this force is provided by the track and is necessary to keep the car moving in a circular path. Without this force, the car would move in a straight line tangent to the loop.