- #1

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I need to find the maximum contortion of a torsion spring, for a given system. Most of the assignment I have done correct, but this last part I simply cannot get right. The pictures below is what I have used, but they are probably not necessary for this problem.

In prior parts of the problem, I figure out, that the angular momentum to the time 0, is

##L=-(l/2+r)*m*v##

Now the ways of writing the torque from the torsion spring, is as follows

##\tau = I*\alpha = -\kappa*\theta##

My thought was, that to find an angular speed to a specific time, when knowing the time, the angular acceleration, and the angular speed at the start, I can use the following:

##\omega=\omega_0+\alpha*t##

I know that my angular momentum, to the specific time where the contortion is at its maximum is 0, as the spring has completely halted the rotational motion. I also know, that this happens at ##t=T/4##, where ##T## is the period (or at least I assume that, as it would make sense thinking about the spring period). As multiplying the ##I## on all factors, gives us an equal that seems desirable, we do that, and get:

$$0=I*\omega_0+I*\alpha*t \Rightarrow 0=-(l/2+r)*m*v+(-\kappa*\theta)*(T/4)$$

From my physics book I also know that ##T=2*\pi*\sqrt{\frac{I}{\kappa}}##. Plugging that in we get

$$0=I*\omega_0+I*\alpha*t \Rightarrow 0=-(l/2+r)*m*v-\frac{\kappa*\theta*\pi*\sqrt{\frac{I}{\kappa}}}{2}$$

Now the ##\theta## here, must to my understanding be the ##\Theta## I'm looking for, as thats that ##\theta## corresponding to the time I have plugged in. Solving for ##\theta## gives me:

$$\theta=\frac{2}{\pi}*\frac{(\frac{l}{2}+r)*m*v}{\sqrt{\kappa*I}}$$

Which according to the solution sheet is correct... except for that ##\frac{2}{\pi}##.

I cannot see what I'm doing wrong, or what I have misunderstood. Please help.

In prior parts of the problem, I figure out, that the angular momentum to the time 0, is

##L=-(l/2+r)*m*v##

Now the ways of writing the torque from the torsion spring, is as follows

##\tau = I*\alpha = -\kappa*\theta##

**The attempt at a solution**

My thought was, that to find an angular speed to a specific time, when knowing the time, the angular acceleration, and the angular speed at the start, I can use the following:

##\omega=\omega_0+\alpha*t##

I know that my angular momentum, to the specific time where the contortion is at its maximum is 0, as the spring has completely halted the rotational motion. I also know, that this happens at ##t=T/4##, where ##T## is the period (or at least I assume that, as it would make sense thinking about the spring period). As multiplying the ##I## on all factors, gives us an equal that seems desirable, we do that, and get:

$$0=I*\omega_0+I*\alpha*t \Rightarrow 0=-(l/2+r)*m*v+(-\kappa*\theta)*(T/4)$$

From my physics book I also know that ##T=2*\pi*\sqrt{\frac{I}{\kappa}}##. Plugging that in we get

$$0=I*\omega_0+I*\alpha*t \Rightarrow 0=-(l/2+r)*m*v-\frac{\kappa*\theta*\pi*\sqrt{\frac{I}{\kappa}}}{2}$$

Now the ##\theta## here, must to my understanding be the ##\Theta## I'm looking for, as thats that ##\theta## corresponding to the time I have plugged in. Solving for ##\theta## gives me:

$$\theta=\frac{2}{\pi}*\frac{(\frac{l}{2}+r)*m*v}{\sqrt{\kappa*I}}$$

Which according to the solution sheet is correct... except for that ##\frac{2}{\pi}##.

I cannot see what I'm doing wrong, or what I have misunderstood. Please help.