A particular unbalanced wheel of a car shakes when the car mov...

Question

A particular unbalanced wheel of a car shakes when the car moves at 90.0 km/h. The wheel plus tire has mass 17.0 kg and diameter 0.58 m. By how much will the springs of this car compress when it is loaded with 280 kg? (Assume the 280 kg is split evenly among all four springs, which are identical.)

Accepted Answer

Welcome back. Everyone in this problem. When a vehicle travels at 80 kilometers per hour, vibrations are noticed due to an imbalanced wheel with a 15 kg mass and a 0.56 m diameter. Calculate the compression of the vehicle's springs when an additional load of 270 kg is evenly placed on the vehicle considering that the vehicle has four identical springs for our answer choices. A says it's 0.0012 m. B 0.0034 m C 0.0043 m and D 0.0062 m. Now, if we're going to find a compression of the vicus springs, let's just make a note of all the information we have so far. So so far we know that the mass of our wheel is 15 kg. Ok. The wheel and the tire. So let's call that m 15 kg. Let's say that the wheel plus the tire, the tires diameter D is equal to 0.50 0.56 m. Sorry, which means that its radius is going to be equal to 0.28 m. The speed that the vehicle is traveling at is 80 kilometers per hour or if we convert that to SI units. Ok. We multiply it by 1000 m divided by 3600 seconds. It would be 22.22 m per second. Eh, and an additional load of 270 kg is placed on the vehicle. Let's call that load capital M so M equals 270 kg. And from all of this, we want to determine the compression of the vehicle springs. In this case, that amount of compression is the amplitude. Now, let's ask ourselves, what do we know that can relate amplitude to all of the information we have so far. Well, recall the equation for elastic force. OK. And recall that equation tells us the elastic force f. In this case, the vehicle's weight MG or the loads weight MG is going to be equal to K multiplied by A where K is the spring constant and A is the amplitude. In that case, then if we're going to solve for a, before we can solve, then we'll need to figure out our value for K. OK. We'll need to eventually solve for K. And when we, once we solve for K, then we'll be able to solve for a. So let's start by thinking about K. Now, since we need the compression of a single tire, we need to divide the vehicle mass by four because we assume the vehicle's weight is distributed equally on all four wheels for all four springs. So we can say then that the mass when we divide it equally is going to be equal to M divided by four. Now, we also know, OK, that our angular velocity is equal to our linear velocity divided by R OK. And we know that it's also equal to the square root of K divided by the mass for or spring. because remember K refers to the spring constant for one of the springs. So this is the mass that we'll be using since both of them are equal to our angular velocity. This means then that the square root of K divided by the mass of the spring is going to be equal to V divided by R I don't know solving for K, we know that we square square, both sides K equals V squared divided by R squared. OK. Sorry K divided by M the mass of the spring. Sorry is gonna be V squared divided by R squared. And remember the mass of the spring is 1/4 of M. So it's gonna be K divided by 1/4 of Mr four K divided by M that's equal to V squared divided by R squared. Let me write that properly here. Now, if we multiply both sides by M and divide both sides by four, that means K equals MV squared divided by four R squared. OK. And no, since we have an expression for K. Yeah, then we'll be able to solve for uh solve for A OK. So now if I put a line here, remember we said MG equals K A, let me put that in red. So that means solving for A by substituting what our value is for K, then MG is going to be equal to that value. Sorry K which is MV squared divided by four R squared multiplied by our amplitude A. OK. Now notice that we can cancel M from both sides. And if we do some transposition here, we'll find that A is going to be equal to four gr squared divided by V squared. Now let's go ahead and substitute our values here. OK? Because no, that means it's going to be equal to four multiplied by G. Let's take our acceleration due to gravity to be 9.8 m per second squared multiplied by our radio squared. So that's 0.28 m squared. OK? And all of that is dividing our velocity squared 22.22 m per second, all squared. Now, when we put that into our calculator to solve, we should finally get a to be equal to 0.0062 m. Thus, that means the compression of the vehicle springs for this additional load is going to be 0.0062 m or from our answer choices. D thanks a lot for watching everyone. I hope this video helped.

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A particular unbalanced wheel of a car shakes when the car moves at 90.0 km/h. The wheel plus tire has mass 17.0 kg and diameter 0.58 m. By how much will the springs of this car compress when it is loaded with 280 kg? (Assume the 280 kg is split evenly among all four springs, which are identical.)

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