The kinetic energy of a particle moving along a circle of radius R depends on the distance covered s as T = as2, where a is a constant. Find the force acting on the particle as a function of s.
The force on the particle at distance s is F = 2as√(1 + (s/R)²). With the numbers on screen, F = N.
Problem statement
Problem 1.81. The kinetic energy of a particle moving along a circle of radius R depends on the distance covered s as T = as2, where a is a constant. Find the force acting on the particle as a function of s.
We assume the particle starts from rest at s = 0 and moves one way round the circle. The distance s is a slider.
We assume m = 1 kg, a = 2 N/m (the unit of a is joule per metre squared, which is newton per metre) and R = 2 m, because the problem gives no numbers. All three are sliders. The mass does not appear in the answer.
Symbols: T is the kinetic energy, v the speed, Ft the force along the path, Fn the force towards the centre, F the total force.
The big idea in plain words
A particle on a circle needs two kinds of force. One speeds it up along the path. The other bends the path towards the centre. The energy rule T = as² tells us how fast the speed grows with distance, so both parts can be found. The two parts are at right angles, so they combine by Pythagoras.
The solution in each frame
Only the ground frame is needed. The circle is at rest in it, and the problem asks for the force in that frame. In a frame turning with the particle, a centrifugal term would appear and cancel Fn. That adds nothing here.
The solution step by step
Step 1, the speed. T = ½mv² = as², so v² = 2as²/m and v = s√(2a/m) = m/s.
Step 2, the force along the path. Differentiate v² = 2as²/m with respect to time, using ds/dt = v. This gives 2v(dv/dt) = 4asv/m, so dv/dt = 2as/m. Then Ft = m(dv/dt) = 2as = N.
Step 3, the force towards the centre. A particle on a circle needs Fn = mv²/R. Put in v² = 2as²/m: Fn = 2as²/R = N.
Step 4, the total force. Ft is along the path and Fn is along the radius, so they are at right angles. F = √(Ft² + Fn²) = 2as√(1 + (s/R)²) = N. It makes an angle with the motion where tan(angle) = Fn/Ft = s/R, which is degrees here.
Step 5, the cases. For s much smaller than R, F is close to 2as and points along the path. For s equal to R the two parts are equal and the angle is 45 degrees. For s much larger than R, F is close to 2as²/R and points almost to the centre.
A separate calculation does not use the force formula. It builds the speed v(s) = √(2as²/m), takes dv/ds by a small step, and finds the acceleration along the path as v(dv/ds) and towards the centre as v²/R. The force is the mass times the size of that acceleration. It gives N, against N from the formula. The relative difference is .
Five methods
1. Force method
Split the acceleration into a path part and a centre part, then add. This is steps 2 to 4.
2. Energy method
The work done by Ft over a small distance ds is dT, so Ft = dT/ds = 2as. This gives the path part at once. The centre part still needs Fn = mv²/R = 2T/R.
3. Graph method
The second graph shows Ft as a straight line and Fn as a parabola. They cross at s = R. The third graph shows the angle of the force, which depends only on s/R.
4. Calculus method
Ft = d(T)/ds = d(as²)/ds = 2as. The speed v = ds/dt gives the same result as differentiating v² in time.
5. Check method
Units: a is in N/m, so 2as is in N. Fn = 2as²/R is N/m times m² over m, which is N. Limits: at s = 0 the force is zero, since the particle is at rest. For R very large the path is straight, Fn goes to zero, and F goes to 2as, the straight-line result.
Fastest: the energy method for Ft, then Fn = 2T/R. Most reliable: the force method with the numerical check.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Mass m | No, for the force | Only the speed v. The force has no m in it |
| Constant a | Yes | The speed, both forces, the total force and the curves |
| Radius R | Yes | The circle, Fn, the total force and the angle |
| Distance s | Yes | The position, the speed, the forces and the picture |
| Given-answer mode | Sets s | The page finds s for a chosen force and warns if s would be above 6 m |
| Play and distance bar | No | Play lets s grow from 0 to the chosen value, holds 1.2 s, then repeats |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Mass | m | kg |
| Constant in T = as² | a | N/m |
| Radius | R | m |
| Distance | s | m |
| Kinetic energy | T = as² | J |
| Speed | v | m/s |
| Force along the path | Ft | N |
| Force towards the centre | Fn | N |
| Total force | F | N |
Exam traps
- Giving only Ft = 2as. A particle on a circle also needs a force towards the centre.
- Giving only Fn = mv²/R. The speed changes, so there is a force along the path too.
- Adding Ft and Fn as numbers. They are at right angles, so use Pythagoras.
- Leaving m in the answer. It cancels, because v² has m in the bottom and the force has it on top.
- Using t as a variable. The problem gives T as a function of distance s, so differentiate with ds/dt = v.