Chapter 4: Problem 14

Determine the time history of the response of the system of Problem \(2.21\) when, over the interval \(0

Short Answer

Expert verified

In this problem, we are asked to find the time history of the response of a system to a given loading function. To do this, we have analyzed the problem under two time intervals, when \(0<t<\tau\) and when \(t\geq\tau\). For each interval, we have derived the response by first solving the homogeneous equation and then finding the particular solution. Finally, we have combined these solutions and applied the necessary initial and boundary conditions to determine the constants \(C_1, C_2, D_1, D_2, A, B, C\). The final expression for the response \(\theta(t)\) for the entire time history is a combination of these solutions, which we can write as: \(\theta(t) = \begin{cases} C_1 e^{r_1t} + C_2 e^{r_2t} + At + B & \text{for } 0<t<\tau\\ D_1 e^{r_1t} + D_2 e^{r_2t} + C & \text{for } t \geq \tau \end{cases}\)

Step by step solution

Find the response when \(0

In this time interval, the flywheel is loaded at a constant rate. We can represent this constant rate by \(M(t) = \frac{M_0}{\tau}t\). In this case, the governing equation of motion is: \(\ddot{\theta}(t) + k\dot{\theta}(t) + g\theta(t) = \frac{M_0}{\tau}t\). To find the response, we will first solve the homogeneous equation and then the particular solution for this interval.

Solve the homogeneous equation for \(0

The homogeneous equation is: \(\ddot{\theta}(t) + k\dot{\theta}(t) + g\theta(t) = 0\). We can solve this equation by assuming a solution of the form \(\theta(t) = e^{rt}\). Substituting this expression into the homogeneous equation gives the following characteristic equation: \(r^2 + kr + g = 0\). Solve this equation for \(r\) to find the two roots.

Find the particular solution for \(0

To find the particular solution, we can use the method of undetermined coefficients. Assuming a linear function \(\theta(t) = At + B\). Differentiating \(\theta(t)\) and \(\dot{\theta}(t)\), we get \(\dot{\theta}(t) = A\) and \(\ddot{\theta}(t) = 0\). Substituting these expressions into the governing equation, we can determine the values of A and B.

Combine the homogeneous and particular solutions for \(0

The combined solution for \(0<t<\tau\) can be written as: \(\theta(t) = C_1e^{r_1t} + C_2e^{r_2t} + At + B\), where \(C_1\) and \(C_2\) are constants that can be determined using initial conditions.

Find the response when \(t\geq\tau\)

In this time interval, the load is maintained at the ultimate level \(M_0\), so \(M(t) = M_0\). The governing equation of motion is: \(\ddot{\theta}(t) + k\dot{\theta}(t) + g\theta(t) = M_0\). Similar to the previous time interval, we will first solve the homogeneous equation and then the particular solution.

Solve the homogeneous equation for \(t\geq\tau\)

The equation is still the same as in Step 2: \(\ddot{\theta}(t) + k\dot{\theta}(t) + g\theta(t) = 0\). Since we already found the roots, we can use the same general homogeneous solution here.

Find the particular solution for \(t\geq\tau\)

To find the particular solution, we can use the method of undetermined coefficients. We assume a constant function \(\theta(t) = C\). Differentiating \(\theta(t)\) and \(\dot{\theta}(t)\), we get \(\dot{\theta}(t) = 0\) and \(\ddot{\theta}(t) = 0\). Substituting these expressions into the governing equation, we can determine the value of C.

Combine the homogeneous and particular solutions for \(t\geq\tau\)

The combined solution for \(t\geq\tau\) can be written as: \(\theta(t) = D_1e^{r_1t} + D_2e^{r_2t} + C\), where \(D_1\) and \(D_2\) are constants that can be determined using boundary conditions at \(t=\tau\).