Crankshaft A crank of radius r rotates with an angular frequen...

Question

Crankshaft A crank of radius r rotates with an angular frequency w It is connected to a piston by a connecting rod of length L (see figure). The acceleration of the piston varies with the position of the crank according to the function

a (Θ) = w²r (cos Θ + (r cos2Θ) / L) .

For fixed w , L, and r find the values of Θ, with 0 ≤ Θ ≤ 2π , for which the acceleration of the piston is a maximum and minimum.

Accepted Answer

Hello, in this video, we're going to be approaching the following word problem. So we are told that a wheel of radius R rotates with an angular speed omega. A point on the edge of the wheel moves in a circular path, and its radial acceleration A of theta varies with its angular position theta according to the function a theta equal to omega 2 R multiplied by 1 minus cosine 2 theta. We want to find the values of theta, with theta being between 0 and 2 pi, for which the radio acceleration of the point is a minimum. OK, so let's go ahead and first write down the given function for the radial acceleration. That function is given to us as A of theta is equal to omega 2 R multiplied by 1 minus cosine of 2 theta. Now, one thing to understand is that the radius and the angular speed are going to be constant in this function. So, in order to determine the values at which this function has minimum values, we need to take the derivative. But by taking the derivative, we just have to keep in mind that omega squared and R are going to be constant values. So, if we take the derivative, we're going to have to use the chain rule. So that will give us a prime of data. is equal to omega 2 R multiplied by the derivative with respect to theta of 1 minus cosine of 2 theta. Now the derivative of 1 is going to be 0. But the derivative of cosine of 2 theta. By the chain rule is going to be negative sign of 2 theta multiplied by the derivative of the innermost function, which is going to be 2. This is going to simplify to be -2 of sin of 2 theta. And since we also have this negative sign in front of cosine 2 theta, the derivative will be positive. So the derivative A of theta is going to simplify to be omega 2 R multiplied by 2 sin of 2 theta, or we can rewrite it as 2 omega 2R multiplied by sin of 2 theta. Now we want to determine the values of theta in which this function is going to have a minimum. So we are going to take the derivative, set it equal to 0, and solve. So in order to solve for theta, we are going to divide both sides of this equation by 2 omega squared R. That is going to leave us with sin of 2 theta equal to 0. And keep in mind we want all the values of theta that are between 0 and 2 pi. Now, the issue here is that we have the angle to theta, so we need to adjust the range of which our solutions can exist. If we take this inequality and multiply each multiply the whole thing by 2, we can rewrite the domain to be 0, less than or equal to 2 theta, which is less than or equal to 4 pi. Now, if we make a substitution and state that Y is equal to 2 theta, we can rewrite the domain and the function as sin of Y is equal to 0 on the interval from 0, less than or equal to Y, less than equal to 4. So now the question becomes, what are the angles of why that we can plug in a sign that caused the sign function to be zero? While those values are going to be wise equal to 0, i, 23, and 4 i. These are all the possible solutions that we get on the interval from 0 to 4 pi. Now, if we resubstitute Y is equal to 2 theta into these solutions, we get 2 theta is equal to 0, i, 23, and 4 i. Originally, we want to solve for all the values of theta that are between 0 and 2 pi. So if we divide this entire expression by 2, we get theta is equal to 0, pi divided by 2, i, 3 pi divided by 2, and 2 pi. Now what we want to do is we want to use the 2nd derivative test to determine which of these values are going to be a minimum. So, just a quick review on the 2nd derivative test. If we plug in a value into the second derivative, and it is greater than 0, this indicates that we have a minimum value. And if we plug in a value into the second derivative, and it is less than 0, this indicates that we have a maximum value. So what we need to do is we need to take the second derivative of this function. By the chain rule, the second derivative, a prime of theta is going to equal. To 4 omega squad are multiplied by cosine of 2 theta. Now, what we are going to do is we are going to plug in all of our possible values of theta into this function and solve. So, if we take these values and plug it into the second derivative, a prime of 0 is going to give us 4 omega squared R, which is greater than 0. If we take the value of pi divided by 2, A prime of pi divided by 2 is equal to -4 omega 2 R, which is less than 0. Plugging in the value of pi is going to give us a prime of pi equal to 4 omega squared R, which is greater than 0. Plugging in 3 pi divided by 2 gives us a of 3 pi divided by 2, which is equal to -4 omega 2 R, which is less than 0. And finally, if we plug in 2 pi into the second derivative, we get 4 omega square R, which is greater than 0. By the second derivative test, this tells us that the minimum values exist at 0. Pie and to pie. And that means that the solutions to this problem is going to be theta is equal to 0, i, and 2 i. And with that being said, the answer to the problem is going to be D. So I hope this video helps you in understanding on how to approach this problem, and I will go ahead and see you all in the next video.

Key Concepts

Angular Motion

Trigonometric Functions

Optimization in Calculus

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