1) A 1500 kg automobile sinks 6cm lower than its normal level when 4 passengers (each of mass 85kg) get in. Determine the frequency of oscillation of the atuomobile with the passengers on board, when it hits a bump on the road. For this problem, I have to first find the spring constant of the spring and then use the formula f=1/(2pi) * square root of (k/m) to calculate the frequency. But is the spring originally uncompressed when no passengers are in? (even though the car has a mass of 1500kg) Or is it already compressed at the beginning? If it's originally compressed, I wouldn't know the displacement from the uncompressed position and therefore I can't calculate the spring constant k. If the spring is uncompressed when no passengers are in, I can say that the new equilibrium position with passengers in is x=-0.06m. (with x=0 representating the unstrecthed position) Fnet=0 -mg+(-kx)=0 -340(9.8)-k(-0.06)=0 k=55533.33N/m But then whe I substitute into f=1/(2pi) * square root of (k/m), which mass should I use, the mass of the car or the passengers? Can someone explain? Thank you! Please Register or Log in to view the hidden image!
It doesn't matter. You don't need to know the initial compression. That's because the spring stretches by the same extra amount every time a 1N weight is added. The mass that the spring has to push and pull around, so I'd say the total mass.
2) A projectile launcher used in the first term has a force constant of 500N/m. A 20g ball is compressed 5cm against the spring and released vertically. 2a) What is the velocity of the ball as it leaves contact with the sprign? 2b) What is the maximum vertical height reached by the ball? 2a) My intuition is that the ball will leave the spring when the spring when the spring is at its unstretched position (i.e. the point where elastic potential energy is zero and has the maximum kinetic energy). But I am not sure if this is true, will it happen this way and why? 2b) Height is relative to the ground, right? So should I assume that the spring is on the ground and is compressed all the way to the ground level? If not, how can I know the height, relative to which point? Can someone explain? Thanks! Please Register or Log in to view the hidden image!
It will. The spring can push the ball up, but it can't pull it back down, so it won't affect the ball after it passes the unstretched position. The ball won't leave the spring until then, either (if it did, the springs upward acceleration would be greater than the ball's, and it'd catch up with the ball). You can only give the final height above the ground if you know the initial height above the ground, which you don't. The best you can do is give the final height above the spring.
2a) How do I know that the ball leaves exactly at that point? If the spring is compressed 5cm, it should spring back up to 5cm stretched, would the ball actually leave the spring at this point instead? And also the ball has some weight, wouldn't the equilibrium position be lower than the unstretched position? (it's like a mass hung on a vertical spring, the equilibrium position is lower than the unstretched position)
The ball would leave at the uncompressed position. If there's a part of the explanation I gave you that you don't understand or agree with, you're welcome to point it out. The spring would keep accelerating the ball (or at least reducing its downward acceleration) past the equilibrium and up to the unstretched position (provided the ball has enough total energy to get there). Yes. You can easily show that the difference is about 40 microns.
Hi, przyk, Please Register or Log in to view the hidden image! But what makes the ball fly out faster than the spring after the uncompressed position? When the spring passes the uncompressed position, it will continue to go up until it's stretched 5cm. I doubt whether the ball will leave when the spring is stretched 5cm, or maybe somewhere in between... Physically, will the ball leave the spring at the equilibrium position or the original uncompressed position? In between these 2 positions, there is still an upward force applied by the spring, I am not sure if this is the reason to say that the ball will stay on a spring until it reaches the uncompressed position... It's getting complex and complicated Please Register or Log in to view the hidden image!
Yo, Simple: the ball isn't attatched to the spring, so the spring can't pull on the ball. At the unstretched position, there's no spring force, so both the spring are accelerating downward at g. Beyond that, the ball is still accelerating downward at g, and the spring is accelerating downward at more than g. Actually it won't get that far. The ball was slowing the spring down while it was going up, so it won't have as much kinetic energy as if the ball had never existed (and all you'd done was compress the spring 5 cm and released it). After the ball flies off, the spring accelerates downward more rapidly due to its lower mass. It is. Think of what would happen if the ball got even one micromillimetre ahead of the spring: The ball's (upward) acceleration would be less than the spring's, and the spring would catch up with the ball. Not really. Try thinking of it this way: The 'natural' thing for the ball to do is to accelerate in the vertical direction at -g (due to gravity). The spring can push on the ball, but it can't pull on the ball. This means that the spring can increase the ball's acceleration, but can't reduce it, right? So the spring and ball will stay together as long as the ball's acceleration is more than -g, but they won't stay together after that.
"At the unstretched position, there's no spring force, so both the spring are accelerating downward at g. Beyond that, the ball is still accelerating downward at g, and the spring is accelerating downward at more than g." Thanks for explaining, it's making more sense now...
