Click into graphic window to stop and restart the simulation. See infos about the Graphic Values and the Experiment Parameters below. If you are new to this, see Explanation of the Cavendish Experiment.
The table below shows turn around and zero tension crossing events.
Get App Url Get App State Set App State Clear Reset Simulation
Use Get App Url to create an url you can post online to call the simulation with the current parameter setting. You can also paste such an url into the field above and press Enter or click on Set App State. This way you can save and restore states for later usage.
The values shown in the graphic are:
Theta θ: is the deflection angle from the zero tension angle α_{0}. It correspond to the red arc in the graphic.
Fg1: is the magnitude of the gravitational force between a fixed and a suspended weight according to Newton's law of universal gravitation. Note, this is not the only force that is acting on the suspended mass and only the tangent component of this force incorporates into the total force of the system.
Fg: is the magnitude of all tangential components of the gravitational forces on the weights m_{2}. If k_{t} = 0 and k_{d} = 0 then this is the only force acting in the system.
Ft: is the tension force due to the torque induced by the string tension constant k_{t}. In equilibrium and if the weights do not touch each other, it is F_{g} = F_{t}.
Change the following values to simulate your configuration:
Speed: defines how much faster the simulations runs than normal speed.
m_{1}, m_{2}: mass of the fixed and suspended weights respectively. See Gravitational Forces how the masses create a gravitational force which moves the suspended weights.
r_{1}, r_{2}: radius of the fixed and suspended weights respectively. The size of the weights has no influence on the movement of the suspended weights. They only define the collision angles.
L_{1}, L_{2}: distances of the fixed and suspended weights from the pivot.
k_{t}: string tension constant that incorporates a torque due to the twisting of the string from the zero tension angle \alpha_0. See Tension Force how this parameter results in a force acting on the suspended weights.
k_{d}: factor to incorporate a linear damping, so that the system does not oscillate infinitely. Real physical systems often have linear and quadratic damping factors. Linear damping means, that the damping force is proportional to the speed. Quadratic damping means, that the damping force is proportional to the square of the speed. Quadratic damping needs only be taken into account for fast moving system. See Drag Force how this parameter results in a force damping the oscillation of the suspended weights.
α_{0}: is the angle of zero tension in the supporting string.
θ_{0}: is the initial deflection angel as measured counter clockwise from α_{0}.
The Cavendish experiment uses two fixed weights and two weights on the ends of a bar which is suspendet on a string so it can rotate around a pivot. This setup cancels the influence of earth's gravity on the experiment. If you place the big fixed weights near the small weights, they attract each other. Because the involved forces are tiny the movement is very slow and it can take a long time to see a result, at least some minutes to hours. Because the gravitational forces between the weights are so tiny, the experiment has to be shielded from air currents and electric and magnetic influences.
The experiment can be used to measure the gravitational constant G if done properly. You can do this by executing the following steps:
To be able to execute steps 3 and 4 you have to design the setup in such a way, that the weights do not touch in equilibrium. The string tension force has to be strong enough to counteract the attraction between the forces. In equilibrium the force due to the string tension is equal the gravitational attraction. So knowing the string tension constant k_{t} and the deflection angle θ due to gravity we can compute the tension force F_{t}, which is equal the gravitational attaction F_{g} of the weights. From the distance of the weights in equilibrium, the measured gravitational forces and the mass of the weights we can compute the gravitational constant G.
To demonstrate that mass attracts mass without quantitatively measure the forces, you can setup the experiment so, that the weights simply collide. In this case you have not to deal with the string tension force, as long as it is not strong enough to hold the weights seprarated.
This App simulates a simple Cavendish experiment setup. It can be used to estimate how long it takes until the gravitational forces between the weights lets them collide. It shows, explains and simulates how the string tension constant can be measured and how the gravitational forces can be obtained from that and how G can be calculated then.
It is assumed that the moving weights are suspended on a string at the center pivot. If twisted, the string can exerpt a torque that counteracts gravity. The magnitude of the torque is dependent on the deflection angle θ of the suspension bar from the zero tension position α_{0} and can be controled by the Parameter k_{t}.
In addition a linear damping force can be applied to the system. The magnitude of the damping is dependent on the angular speed of the suspension bar and can be controled by the Parameter k_{d}.
If you create the Canvendish experiment yourself, you probably don't know the string tension constant k_{t}. But you can measure it using the following method:
(Jump directly to the formula (5) if you have already measured T to compute k_{t})
The idea behind this is: The string acts like a spring and the suspension bar like a mass. So we have a springmass system that can oscillate. The tension constant corresponds to the spring constant. The only difference is, that we have a rotating system, where the tension constant gives a torque rather than a force and the mass is replaced by a moment of inertia.
The oscillation period of a massspring system can be computed as follows:
(1) 
 
where^{'} 

Note that the period is measured from one point of oscillation until it reaches the same point from the same direction.
Tip: If you have multiple time periods measured in one run, average over the last few periods, because the influence of damping and nonlinearities in the system is smallest when the amplitude of the oscillation is smallest.
The oscillation period of a stringsuspension bar system can be computed very similarly:
(2) 
 
where^{'} 

If we have measured the oscillation period T, we can compute the corresponding string tension constant by solving (2) for k_{t}:
(3) 
 
where^{'} 

But what is this Moment of inertia I? It is a quantity that determines the torque needed for a desired angular acceleration about a rotational axis; similar to how mass determines the force needed for a desired linear acceleration.
The moment of inertia depends on the mass distribution of an object. Is can be computed by dividing an object into many small volume elements m_{i}. The moment of inertia of each element is its mass times the square of its distance from the rotation axes: I_{i} = m_{i} · r_{i}^{2}.
For a suspension bar with heavy weights at the end, so that the weight of the bar is negligible, like in the simulation, the moment of inertia is simply:
(4) 
 
where^{'} 

See Computing the Moment of Inertia for the Suspension Bar, if the mass of the bar itself is not neglegtable or if you want higher precision measurements.
So for the suspension bar system in the simulation or for a corresponding real model we can compute the string tension constant by inserting (4) into (3):
(5) 
 
where^{'} 

Set Simulation State for this Example
To measure the string tension constant we remove the fixed weights. Then we displace the suspension bar from the zero tension angle (the rest position α_{0}) some degrees eg. θ_{0} = 10° and let the suspension bar oszillate. We stop the time T of one full oszillation.
Note: due to damping and some nonlinearities in the suspension string the most accurate measurement for T is when the oszillation angle is as small as possible. But this can make it difficult to stop the time accurately. One way to get better results is to average over some periods.
My simulation got me a period T = 6562 s (from the event table below the simulator).
If the weight of the suspension bar is neglegtable with respect to the weights m_{2}, I can directly use formula (5) to calculate the string tension constant k_{t}:
(6) 
 
where^{'} 

which is very close to the value k_{t} = 20 × 10^{−9} N·m/rad that I entered in the simulation for this example.
To measure the string tension constant k_{t} with the formula (3) you need the moment of inertia I of the suspension bar with the weights m_{2}. If the weight of the bar itself is much smaller than the weights at the end, you can neglegt the moment of inertia of the bar and simply use (5) to compute k_{t}.
If the mass of the bar is not neglegtable or if you want higher precision measurements, you can incorporate the moment of inertia of the bar itself. Assuming the bar is a rod with any cross section shape, but constant diameter and density, you can compute its moment of inertia as follows:
(7) 
 
where^{'} 

The total moment of inertia of the suspension bar inclusive the weights at the end is then:
(8) 
 
where^{'} 

Note that L_{bar} is the length of the bar without the weights. So dependent of your construction L_{bar} may be smaller or bigger than 2 · L2.
To measure the gravitational force it is not sufficient to simply let the weights come together. You have to setup the system in such a way, that the weights come closer but never touch. Wait until the oscillation stops. Now you can measure the angle θ from the zero tension position α_{0} to the current position. In this equilibrium state the momentum due to the string tension is proportional to the gravitational force.
If we now the string tension constant k_{t} and the angle θ we can compute the torque and from that the corresponding torque forces at the positions of the 2 weights m_{2}. In equilibrium the sum of this forces is just equal to the total gravitational force:
(9) 
 
where^{'} 

Now we have the sum of all tangential acting components of the gravitational forces F_{g} as displayed in the simulator. The force acting at one mass is half of that.
To compute the gravitational constant we need not the total tangential components of the gravitational forces but the force vector from one mass m_{2} to the nearest mass m_{1}. We have essentially to reverse engineer what the simulation does as described at Gravitational Forces to get \vec F_{g1}. This is a tedious calculation and shown at (11) for completeness.
But we can use F_{g} as an approximation:
(10) 
F_{g1} \approx { F_g \over 2 }

To get the real force \vec F_{g1} between two weights m_{1} and m_{2} we have to reconstruct the force vectors from the tangential components. For that we need the geometry of the experiment at the stabilized state. The following used geometric parameters are defined under Gravitational Forces.
The real force is then:
(11) 
 
with 
\hat d_1 = { \vec D_1 \over d_1 } \qquad\quad \hat d_3 = { \vec D_3 \over d_3 } \qquad\quad d_1 =  \vec D_1  \qquad\quad d_3 =  \vec D_3 
 
and 
\vec D_1 = \vec P_1  \vec P_2 \qquad\quad \vec D_3 = \vec P_3  \vec P_2 \qquad\quad \hat r_t = \left\lgroup \matrix{ \sin\left( \alpha_0  \theta \right) \\ \cos\left( \alpha_0  \theta \right) } \right\rgroup

The formula to compute a gravitational force between two weights is:
(12) 
F = { G \cdot m_1 \cdot m_2 \over d^2 }

If we know F, the masses and the distance d between the weights, we can calculate the gravitational constant G by solving (12) for G:
(13) 

F is our F_{g} / 2 from Measuring the Gravitational Force.
To check that the formulas are correct, you can use the force displayed as Fg1 in the simulation. This is the pure force between two weights m_{1} and m_{2} (not their tangential component).
What we still don't know is the distance d between the weights. We can measure it or calculate it from the deflection angle θ:
(14) 
d = \sqrt{ {L_1}^2 + {L_2}^2  2 \cdot L_1 \cdot L_2 \cdot \cos( \alpha_0  \theta ) }

Set Simulation State for this Example
In this example I got the following deflection angle after some hours after the suspension bar stoped oscillating: θ = 2.411°. Assuming we have already measured the string tension constant k_{t} = 20 nN·m/rad, we can compute the tension force which is the same as the total gravitational forces:
(15) 
 
where^{'} 

Lets calculate the distance between the weights:
(16) 
 
where^{'} 

The gravitational constant using the measured force F ≈ F_{g} / 2 is then:
(17) 
 
where^{'} 

If we use the exact gravitational force as displayed at Fg1 and computable with (11) we get:
(18) 
 
where^{'} 

which is the publised value.
The simulation is not an algorithmus to produce an animation. I tought the computer how physics works. He computes all forces from the current state, taking all parameters into account. The sum of all forces results in the acceleration of the suspended weights. Numerical Integration of the acceleration delivers the next state after a certain small time interval: the new rotation speed and position. From the new state all forces are computed again, and so forth.
To calculate the gravitational forces we need to define some points as vectors, for which the origin is at the pivot of the weights m_{2}:
(19) 
 
where^{'} 

Furthermore we need the vectors from \vec P_2 to \vec P_1 and \vec P3 and their lengths:
(20) 
\vec D_1 = \vec P_1  \vec P_2
 
(21) 
d_1 =  \vec D_1  = \sqrt{ { ( { D }_{ 1,x } ) }^2 + { ( { D }_{ 1,y } ) }^2 }
 
(22) 
\vec D_3 = \vec P_3  \vec P_2
 
(23) 
d_3 =  \vec D_3  = \sqrt{ { ( { D }_{ 3,x } ) }^2 + { ( { D }_{ 3,y } ) }^2 }
 
where^{'} 

The gravitational force vectors are then:
(24) 
 
where^{'} 

Note: the term \vec D_1 / d_1 defines a unit vector from P_{2} to P_{1}, which gives the direction of the force \vec F_{g1} due to mass P_{1}. The same applies to \vec F_{g3}.
Because the suspended weights are not free to move in any direction but are constrained to rotate around the pivot point, only the tangent components are relevant for the movement. The radial components of the forces cancel each other. To calculate the tangent components we introduce a tangential unit vector \hat r_t as shown in the image above:
(25) 
\hat r_t = \left\lgroup \matrix{ \sin\left( \alpha \right) \\ \cos\left( \alpha \right) } \right\rgroup

Now we can calculate the tangential components of the gravitational forces by applying the vector dot product:
(26) 
 
where^{'} 

To get the total gravitational force we add the tangential force components and multiply by 2 because we have 2 suspended weights:
(27) 
 
where^{'} 

The sum of the tangent components of the gravitational forces is displayed as Fg in the simulation.
Note: F_{g3,t} points in the opposite direction of F_{g1,t}, which is taken into account by an opposite sign, resulting from the vector dot product with \hat r_t.
The string exerts a torque on the suspended mass system that is proptional to the rotation angle measured from the zero tension angle \alpha_0. The torque acts always in the direction to the zero torque angle. To compute the sum of all forces, we need the torque in terms of a tension force acting at the locations of the weights m_{2}.
The magnitude of the torque is:
(28) 
M = F_t \cdot L_2 = k_t \cdot ( \alpha  \alpha_0 )

The torque is acting against the displacement from α_{0}, so the tension force is minus:
(29) 
 
where^{'} 

Note: this is the sum of the two tension forces acting at each mass m_{2}.
If the tension constant k_{t} is too big, the gravitaional forces can not move the weights together, but only deflect them in the corresponding direction. The tension force is displayed as Ft in the simulation.
To introduce a damping of the system, we can define a linear damping constant k_{d}, which produces a tangential damping force always acting against the direction of motion. The magnitude of the damping force is proportional to the angular speed.
(30) 
 
where^{'} 

If you set k_{d} = 0 the system is undamped and oscillates indefinitely.
Now we get to the equations of motion. If we know all forces acting on the suspended part of the system, we can compute the angular acceleration applying Newton's second law of motion a = F / m:
(31) 
 
where^{'} 

Note: to get from the linear acceleration to the angular acceleration, we had to divide by L_{2}.
It is common practice to write accelerations with two dots and speeds with one dot above the angle variable. One dot means a differentiation with respect to time. Two dots mean two times differentiated with respect to time.
From the angular acceleration we can calculate the angular speed by integration:
(32) 
 
where^{'} 

From the angular speed we get the angel by integration:
(33) 
 
where^{'} 

The integration can not be done analytically for such complicated systems, but have to be done numerically by a computer program.
In principle we divide the motions into small chunks of time length \Delta t. Then we compute for each time step from the known acceleration angle the change in angular velocity and the change in angular position. We add the change in angular velocity to the current angular velocity and the change in angular position to the current angular position and get angular velocity and position for the next time step:
(34) 
\Delta \dot \alpha( t ) = \ddot \alpha( t ) \cdot \Delta t

\dot \alpha( t + \Delta t ) = \dot \alpha( t ) + \Delta \dot \alpha( t )

(35) 
\Delta \alpha( t ) = \dot \alpha( t ) \cdot \Delta t

\alpha( t + \Delta t ) = \alpha( t ) + \Delta \alpha( t )

This steps are continuously repeatet for each time step \Delta t and the objects are drawn at the calculated positions.
Note: the simulation uses a more sophisticated rule for the integration, that halves the error compared to the above method.
The weights can collide if the difference of L_{1} and L_{2} is smaller than the sum of the radii r_{1} + r_{2} of the weights.
If the weights can not collide, nothing has to be done, the simulation just works. But if the weights can collide, their angular speed has to be reversed on collision detection and the angular position has to be corrected.
We assume an elastic collision. That means the momentum is conserved without energy loss on the collision. This is established by simply inverting the angular speed on collisions.
To check for collisions we need the minimum and maximum angular positions of the suspension, that correspond to the collision positions. We can get the angles geometrically from the triangle P_{1}, Pivot, P_{2} when the weights touch each other. In this case P_{1} and P_{2} are at a distance of r_{1} + r_{2} and the other sides of the triangle are L_{1} and L_{2}:
(36) 
 
(37) 
\alpha_\mathrm{max} = \pi  \alpha_\mathrm{min}
 
where^{'} 

A minor point regarding the moment of inertia of the setup. Your formula (4)
would only hold if the balls itself were non rotating. Otherwise you need to add the moment inertia of the balls rotating around their center of mass, which for each individual ball is \frac{2}{5} m_2 {r_2}^2. So that gives
in total, according to Steiner's theorem. Note that this only holds if the balls are massive. Hollow balls have a different moment of inertia.