- #1

TimmyD1

- 10

- 1

- Homework Statement:
- System of spring and pendulum

- Relevant Equations:
- L=T-V

Hello! I have some problem getting the correct answer for (b).

*My FBD:*$$L=T-V\iff L=\frac{1}{2}m(b\dot{\theta})^2+mg(b-b\cos\theta)-\frac{1}{2}k\boldsymbol{x}^2,\ where\ \boldsymbol{x}=\sqrt{(1.25b-b)^2+(b\sin\theta)^2}-(1.25b-0.25b)$$

Hence my equation of motion become

$$\frac{1}{2}mb^2\ddot{\theta}+mg(b-b\cos\theta)-\frac{1}{2}k\boldsymbol{x}^2=0$$

solving this gives me the spring coefficient $$k=111.92\ N/m$$.

for part (b):

when $$\theta=25^{\circ}\ and\ k=100\ N/m$$ then

$$\frac{1}{2}mb^2\dot{\theta}\frac{d\dot{\theta}}{d\theta}+mg(b-b\cos\theta)-\frac{1}{2}k\boldsymbol{x}^2=0\iff mb^2\dot{\theta}\frac{d\dot{\theta}}{d\theta}+2mgb(1-\cos\theta)-k\boldsymbol{x}^2=0\iff \dot{\theta} d\dot{\theta} = \Big[\frac{k\boldsymbol{x}^2}{mb^2}-\frac{2g}{b}(1-\cos\theta)\Big]\ d\theta$$

$$\int_0^{\dot{\theta}} \dot{\theta} d\dot{\theta} = \int_0^{\frac{25\pi}{180}}\underbrace{\Big[\frac{k\boldsymbol{x}^2}{mb^2}-\frac{2g}{b}(1-\cos\theta)\Big]}_{-1.7001}\ d\theta\Longrightarrow \frac{\dot{\theta}^2}{2}=0.74384\iff \dot{\theta}=\sqrt{2\cdot 0.74384}=1.2197\ rad/s\Longrightarrow v=r\omega=b\dot{\theta}=0.4\cdot1.2197\approx 0.48788\ m/s$$

The answer should be

**v = 0.522 m/s**but I can't seem to get it right, would really appreciate it if anyone could help me out!