# Calculating velocity of two different weight in a system with spring

I get a question like this. But what if the angle is given as theta=60° and I'm required to find the velocity of each weight. I tried the work energy equation but it doesn't seem to work since there are two unknown.

The second picture are my attempt at it. I'm just stuck since there are two unknown and no other equation to relate both velocity to each other.

• It looks like howework so you need to show your work and where you are stuck. Commented Dec 23, 2023 at 13:59
• I have edited the question to insert my work Commented Dec 23, 2023 at 15:21

The total energy of the system is

$$T=4 g\ \frac{2(200)}{1000}+2\ (3 g)\ \frac{ 500}{1000}=\frac{23 \ g}{5}=\frac{1127}{25}$$

At a value $$\theta$$, the position of the $$4 \ kg$$ mass is

$$\{p_{x1},p_{y1}\}=\{0,\frac{2(200\ \sin (\theta/2))}{1000}\}=\{0,\frac{2}{5} \sin \left(\frac{\theta }{2}\right)\}$$

And its velocity is

$$\{v_{x1},v_{y1}\}=\{0,\frac{1}{5} \dot{\theta } \cos \left(\frac{\theta }{2}\right)\}$$

The position of the $$3 \ kg$$ mass is

$$\{p_{x2},p_{y2}\}=\{\frac{200 \cos \left(\frac{\theta }{2}\right)+300 \cos \left(\frac{\theta }{2}\right)}{1000},\frac{200 \sin \left(\frac{\theta }{2}\right)+300 \sin \left(\frac{\theta }{2}\right)}{1000}=\{\frac{1}{2} \cos \left(\frac{\theta }{2}\right),\frac{1}{2} \sin \left(\frac{\theta }{2}\right)\}\}$$

And its velocity is

$$\{v_{x2},v_{y2}\}=\{-\frac{1}{4} \dot{\theta } \sin \left(\frac{\theta }{2}\right),\frac{1}{4} \dot{\theta } \cos \left(\frac{\theta }{2}\right)\}\}$$

The kinetic energy at a value $$\theta$$ is

$$K=\frac{1}{2} 4\ v_{y1}^2+2\frac{1}{2} 3\ (v_{x2}+v_{y2})^2\\ \ \ \ \ =\frac{2}{25} \dot{\theta }^2 \cos ^2\left(\frac{\theta }{2}\right)+\frac{3}{16}\dot{\theta }^2$$

The potential energy at a value $$\theta$$ is

$$U=4\ g\ p_{y1}+2 \ (3\ g)\ p_{y2}+\frac{1}{2} k(\frac{400}{1000}-p_{y1})^2\\ \ \ \ =450 \left(\frac{2}{5}-\frac{2}{5} \sin \left(\frac{\theta }{2}\right)\right)^2+\frac{1127}{25} \sin \left(\frac{\theta }{2}\right)$$

By the law of conservation of energy

$$\fbox{T=K+U}$$

Case 1: $$\theta=60^{\circ}$$

$$\frac{1127}{25}=\frac{99 \ \dot{\theta }^2}{400}+\frac{2027}{50}$$

The negative solution (for downward motion) is $$\dot{\theta }=-\frac{2}{3}\sqrt{\frac{454}{11}}$$

Substituting $$\theta=60^{\circ}$$ and $$\dot{\theta }=-\frac{2}{3}\sqrt{\frac{454}{11}}$$ in $$\{v_{x1},v_{y1}\}$$ and $$\{v_{x2},v_{y2}\}$$ we get

$$\{v_{x1},v_{y1}\}=\{0, -\frac{1}{5}\sqrt{\frac{454}{33}}\}$$

$$\{v_{x2},v_{y2}\}=\{\frac{1}{6}\sqrt{\frac{227}{22}},-\frac{1}{2}\sqrt{\frac{227}{66}}\}$$

Case 2: $$\dot{\theta}=0$$ (the lowest position)

$$\frac{1127}{25} = 450 \left(\frac{2}{5}-\frac{2}{5} \sin \left(\frac{\theta }{2}\right)\right)^2+\frac{1127}{25} \sin \left(\frac{\theta }{2}\right)$$

To solve, first set $$p=\sin \left(\frac{\theta }{2}\right)$$ to get

$$450 \left(\frac{2}{5}-\frac{2 \ p}{5}\right)^2+\frac{1127 \ p}{25}=\frac{1127}{25}$$

This quadratic has two solutions $$p=\frac{673}{1800}$$ and $$p=1$$. The latter gives $$\theta=180^{\circ}$$ which is the topmost position. We use the other value $$p=\frac{673}{1800}$$ to get the $$\theta$$ for the lowest position

$$\theta\ =\ 2\ \sin ^{-1}\left(\frac{673}{1800}\right)$$

Which turns out to be around $$44 {}^{\circ}$$.

Bonus You can also visualize how the kinetic and potential energies are traded during the downward motion. The system starts with $$0$$ K.E. and it reaches a max at around $$87^{\circ}$$ and goes back to $$0$$. Conversely, the P.E. is a minimum at around $$87^{\circ}$$.

I think your trouble starts here:

Figure 1. The wrong turn?

I don't think you should be looking at work, but look at force instead.

We can simplify the problem by doing a pre-calculation for the effect of the weight:

Figure 2. The effect of the weight on the compressed spring.

• I agree that the spring will have compressed 0.2 m so its reaction will be 900 × 0.2 = 180 N as shown.
• The 4 kg mass will account for 40 N of that. (I'm old so I don't have time to mess around with calculating a more exact value.)
• That means that the weights are contributing 140 N or 70 N each in the vertical direction.

If you agree with that, then please recalculate and post your updated solution under the original. (Leave the original so that my answer doesn't look stupid.)

• I think this is a simple harmonic motion problem, hence the maximum compression is when the potential energy of the masses has been transferred to elastic energy in the spring. Commented Dec 23, 2023 at 19:01
• @GregLocock, how is it simple harmonic motion? I'm only familiar with that in the case of a pendulum but can't see how that's relevant here. And I don't see any potential energy here. Commented Dec 23, 2023 at 20:06
• OK, what will happen after the springs are fully compressed? The weights will bounce up, back to their original position, and then back down ad infinitum in the absence of damping and friction. Commented Dec 23, 2023 at 20:20
• @GregLocock, this is a speed governor similar to those used on steam engines or stationary diesels. The throttle is wide open initially but as the balls are thrown out by rotary motion the spring is compressed, the throttle closes down and it should stabilise at the set speed. If the load increases the engine will slow down reducing the outward force on the weights which can then be pushed back up by the spring, and opening the throttle some more ... Commented Dec 23, 2023 at 20:33
• In that case, wouldn't the system be in equilibrium since all the force will cancel each other out. That's not to mention that the weight on the right and left side are both constant 3kg and only experience gravitational acceleration. Of course that's unless the force from the spring push the block upward since it has the bigger force Commented Dec 24, 2023 at 1:52

## Conservation of mechanical energy

In this exercise you can use the principle of conservation of mechanical energy because there are conservative forces only, and since you have a starting position $$(\theta=\pi, \dot\theta=0)$$ and a given value of the angle $$\theta=\frac{\pi}{3}$$ you get:

$$K(\pi) + U(\pi) = K(\frac{\pi}{3}) + U(\frac{\pi}{3})$$

but since the velocity of each part is zero at the starting position, this simplifies to:

$$U(\pi) = K(\frac{\pi}{3}) + U(\frac{\pi}{3})$$

and since this is a 2-dimensional problem on a plane, we have:

$$K =\sum\limits_{i} \left[ \frac{1}{2}m_i(\dot x_i^2+\dot y_i^2) + \frac{1}{2}I_i\dot\theta_i^2\right]$$
$$U = \sum\limits_{i}\left[ m_igy_i+\frac{1}{2}k_i(\Delta l_i)^2 \right]$$

## Writing the coordinates w.r.t. $$\theta$$

This mechanism only has one degree of freedom $$(\theta)$$ so each unknown in the equations above can be specified w.r.t. said variable.

For the mass in the center:

$$y_M=2a\sin(\frac{\theta}{2})$$
$$\dot y_M=a\cos(\frac{\theta}{2})\dot\theta$$
$$\dot y_M^2=a^2\cos^2(\frac{\theta}{2})\dot\theta^2$$

Here I just derived the position to get the modulus of the velocity squared because the mass is constrained to move in a vertical line.

For the spherical masses it is more convenient to express their kinetic energy through the moment of inertia $$I_m=m(a+b)^2$$ and the angular velocity because they are pivoting about a fixed point at the base of the mechanism. For the gravitational potential energy you still need their y-coord however. We find both:

$$y_m=(a+b)\sin(\frac{\theta}{2})$$
$$\theta_m=\frac{\theta}{2}$$
$$\dot\theta_m=\frac{\dot\theta}{2}$$
$$\dot\theta_m^2=\frac{\dot\theta^2}{4}$$

For the spring we have:

$$\Delta l= l(\theta) - l_{rest}=y_M-2a=2a\left(\sin\left(\frac{\theta}{2}\right)-1\right)$$
$$(\Delta l)^2=4a^2\left(\sin\left(\frac{\theta}{2}\right)-1\right)^2$$

where, respectively:

$$a=0.2\, m$$
$$b=0.3\, m$$
$$m=3\, kg$$
$$M=4\, kg$$
$$k=900\, N/m$$

$$a$$ and $$b$$ are the lengths of the two portions of the rods holding the spherical masses, $$M$$ and $$m$$ are the central mass and the spherical masses respectively.

## Particularizing the energies

Now we just plug into $$K(\theta)$$ and $$U(\theta)$$ the expressions we just found. For the kinetic energy we have:

\begin{align} K(\theta)&=\frac{1}{2}M\dot y_M^2+2\cdot \frac{1}{2}I_m\dot\theta_m^2\\ &=\frac{1}{2}Ma^2\cos^2\left(\frac{\theta}{2}\right)\dot\theta^2+m(a+b)^2\frac{\dot\theta^2}{4}\\ &=\left[4Ma^2\cos^2\left(\frac{\theta}{2}\right)+2m(a+b)^2\right]\frac{\dot\theta^2}{8} \end{align}

And the values we're interested in are:

$$K(\pi)=0$$
$$K(\frac{\pi}{3})=\left[3Ma^2+2m(a+b)^2\right]\frac{\dot\theta^2}{8}$$

The potential energy is:

\begin{align} U(\theta)&=Mgy_M+2\cdot mgy_m+\frac{1}{2}k(\Delta l)^2\\ &=2Mga\sin\left(\frac{\theta}{2}\right)+2mg(a+b)\sin\left(\frac{\theta}{2}\right)+2ka^2\left(\sin\left(\frac{\theta}{2}\right)-1\right)^2 \end{align}

And again, the values we need are:

$$U(\pi)=2Mga+2mg(a+b)$$
$$U(\frac{\pi}{3})=Mga+mg(a+b)+\frac{ka^2}{2}$$

Now we substitute these values in the mechanical energy conservation formula and get:

$$2Mga+2mg(a+b)=\left[3Ma^2+2m(a+b)^2\right]\frac{\dot\theta^2}{8}+Mga+mg(a+b)+\frac{ka^2}{2}$$

After simplifying you can get the angular velocity as:

$$\dot\theta\left(\frac{\pi}{3}\right)=\sqrt{\frac{2Mga+2mg(a+b)-ka^2}{\frac{3}{4}Ma^2+\frac{1}{2}m(a+b)^2}}$$

which will now be referred simply as $$\dot\theta$$ to keep notation clear and simple. Another detail is that this is the absolute value of the angular velocity.

## Finding the velocities

Now we can find the velocities of each mass by using the relations found earlier which contain both $$\theta$$ and $$\dot\theta$$.

For the velocity of the central mass we simply have again:

$$\dot y_M=a\cos(\frac{\theta}{2})\dot\theta$$

and we have to remember that the mass is travelling downwards, so this will be a negative velocity.

For the two spherical masses we have that their velocity is calculated using the formula of circular motion:

$$|v|=\omega r$$

where, in our case, $$\omega=\dot\theta$$ is the angular velocity and $$r=a+b$$ is the distance from pivot to mass, so we get:

$$|v_m|=\dot\theta(a+b)$$

If you need the components in the x-y axes, you can calculate them, knowing that the angle between the ground and the arm holding the spherical masses is $$\frac{\theta}{2}=\frac{\pi}{6}$$, so with a bit of trig you get:

$$\dot x_{m2}=-\dot x_{m1}=|v|\sin\left(\frac{\pi}{6}\right)$$
$$\dot y_{m1}=-|v|\cos\left(\frac{\pi}{6}\right)$$

where $$m_1$$ is the spherical mass on the left, $$m_2$$ is the one on the right.

• Since the weights are stationary at maximum compression. which is all the OP requires, all the impressive equations for velocity seem a bit redundant. Commented Dec 25, 2023 at 0:05
• Op asks for the velocity of each weight at an angle of 60° despite what the provided picture says, so I provided such result. Commented Dec 25, 2023 at 10:10
• You are absolutely right, I read the printed bit! Commented Dec 25, 2023 at 21:37