In the arrangement shown in Fig. 1.17, the mass of the rod M exceeds the mass m of the ball. The ball has an opening permitting it to slide along the thread with some friction. The mass of the pulley and the friction in its axle are negligible. At the initial moment, the ball was located opposite the lower end of the rod. When set free, both bodies began moving with constant accelerations. Find the friction force between the ball and the thread if t seconds after the beginning of motion the ball got opposite the upper end of the rod. The rod length equals l.
The friction force between the ball and the thread is F = 2lmM/((M − m)t²). With the numbers on screen, F = N. The ball then moves .
Problem statement
Problem 1.73. In the arrangement shown in Fig. 1.17, the mass of the rod M exceeds the mass m of the ball. The ball has an opening permitting it to slide along the thread with some friction. The mass of the pulley and the friction in its axle are negligible. At the initial moment, the ball was located opposite the lower end of the rod. When set free, both bodies began moving with constant accelerations. Find the friction force between the ball and the thread if t seconds after the beginning of motion the ball got opposite the upper end of the rod. The rod length equals l.
Interpretation. A rod of mass M hangs from a thread that runs over a fixed pulley. The mass M of the rod is more than the mass m of a ball. The ball has an opening that lets it slide along the other part of the thread, with some friction. The pulley has no mass and its axle has no friction. At the start the ball is level with the lower end of the rod. When set free, both bodies begin moving with constant accelerations. Find the friction force between the ball and the thread, if t seconds after the start the ball is level with the upper end of the rod. The rod has length l.
We assume m = 0.1 kg, M = 0.4 kg, l = 1 m and t = 0.45 s to start. The problem uses letters only. All four are sliders.
We assume g = 9.8 m/s². It is a slider.
We assume the friction force F stays constant while the ball slides, the thread is long enough that the ball never reaches its end, and the thread has no mass.
Symbols: w₁ is the downward acceleration of the rod, w₂ the downward acceleration of the ball, T the tension in the thread at the rod, and F the friction force between the ball and the thread.
The big idea in plain words
If there were no friction, the thread would have no tension and both bodies would fall freely together. The ball would stay level with the lower end of the rod for ever. The friction is what makes the difference.
The ball drags the thread down with the force F. The thread has no mass, so it passes this same force on to the rod, pulling it up with T = F. The ball is lighter than the rod, so the same force F slows the ball more than it slows the rod. The rod falls faster and the ball ends up level with the upper end of the rod.
The solution in each frame
Ground frame: write Newton's second law for the rod and the ball (steps 1 to 3). Frame of the rod: this frame accelerates, but a pseudo-force acts on the rod and the ball in the same way (m and M times the same acceleration), and it cancels in the relative acceleration. So in the rod's frame the ball simply starts from rest and rises with the relative acceleration w₁ − w₂, which gives the climb equation in step 4. The picture follows the rod.
The solution step by step
Step 1, the force on the thread. The ball pulls the thread down with the friction force F. The thread has no mass, so the tension at the rod is T = F.
Step 2, the rod. Take down as positive. Gravity pulls it down with Mg, and the thread pulls it up with F. So Mg − F = M w₁, and w₁ = g − F/M = m/s².
Step 3, the ball. Gravity pulls it down with mg, and the friction from the thread pushes it up with F. So mg − F = m w₂, and w₂ = g − F/m = m/s².
Step 4, the relative acceleration. w₁ − w₂ = F(1/m − 1/M) = F(M − m)/(mM), which is positive because M is more than m. So the rod falls faster than the ball. In the frame of the rod the ball starts from rest and rises with this acceleration. It reaches the upper end after a rise of l: l = ½(w₁ − w₂)t².
Step 5, the answer. Put step 4 into the climb equation: l = ½ F(M − m)t²/(mM). So F = 2lmM/((M − m)t²) = N.
Step 6, what is possible. The ball slides down along the thread only if its acceleration downwards is bigger than the downward acceleration of the thread, which is minus w₁ (the thread moves up as the rod goes down). So w₂ + w₁ = 2g − F(1/m + 1/M) must be above zero. This means F must not be above 2gmM/(M + m) = N. In terms of time, t must not be below √(l(M + m)/(g(M − m))) = s. At that shortest time the ball does not slip at all: it is fixed on the thread and the system is an ordinary two-body pulley.
Step 7, the cases. The ball's acceleration w₂ = g − F/m changes sign at F = mg = N.
- Case a, F < mg: the ball falls, but more slowly than the rod.
- Case b, F = mg: the ball hangs still in the air while the rod falls past it. This happens at t = √(2lM/((M − m)g)) = s.
- Case c, F > mg: the ball is carried upwards while the rod falls. This is possible down to the shortest time above.
The rod itself always falls, because F is never as large as Mg (F is at most 2gmM/(M + m), which is below Mg).
A separate calculation solves the four equations (rod, ball, tension and the climb condition for the given time) as one linear system. It gives F = N, against N from the formula. It then lets the ball climb on the rod in 400000 small time steps, using that force, and finds that it reaches the upper end after s, against t = s. The largest relative difference is .
Five methods
1. Force method
Newton's second law for the rod and the ball, then the climb equation. This is steps 1 to 5.
2. Energy method
The friction force does work F times the sliding distance, and this is turned into heat. The sliding distance along the thread is ½(w₁ + w₂)t², which needs w₁ and w₂ first. The energy method therefore adds nothing here beyond a check.
3. Graph method
The third graph plots the height of the rod and of the ball against time, with falling drawn downwards. Both start together. The rod drops faster, so the gap between the curves grows as ½(w₁ − w₂)τ², and it equals the rod length l at τ = t. That gap is the climb of the ball on the rod.
4. Calculus method
Because F is constant, both accelerations are constant. Integrate the relative acceleration twice with zero starting speed and position to get ½(w₁ − w₂)t².
5. Check method
Units: l m M/(M − m) divided by t² is kg m/s², which is a force. Limits: for a very long time, F tends to zero and the bodies fall freely. If M is only a little above m, F becomes large for the same t, since the two bodies are then hard to separate. At the shortest time, F = 2gmM/(M + m), the tension of an ordinary two-body pulley.
Fastest: the force method with the relative acceleration. Most reliable: the force method followed by the numerical check, because it also tests the range of times that are possible.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Mass of the ball m | Yes | F, both accelerations, the ball size in the picture and every curve |
| Mass of the rod M | Yes | F, both accelerations, the rod width in the picture and every curve |
| Rod length l | Yes | The rod in the picture, F, the climb of the ball and the curves |
| Time t | Yes | F, both accelerations, the run time and the curves |
| Gravity g | Yes | F and both accelerations |
| Case buttons a, b, c | Set t | a: t = 1.5 times the time of case b. b: F = mg. c: halfway between the shortest time and case b |
| Run a, b, c | No | Plays the three cases in order |
| Given-answer mode | Sets t | The page finds the time for a chosen F, and warns if F is too large |
| Play and time bar | No | Play runs the motion at half speed (faster for long times), holds 1.2 s at the end, then repeats. The bar sets the time |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Mass of the ball | m | kg |
| Mass of the rod | M | kg |
| Length of the rod | l | m |
| Gravity | g | m/s² |
| Time | t | s |
| Friction force | F | N |
| Accelerations of the rod and the ball | w₁, w₂ | , m/s² |
| Largest possible force | 2gmM/(M + m) | N |
Exam traps
- Treating the thread as fixed, so that the ball moves up the thread by l. The thread itself moves. Only the ball and the rod are compared.
- Forgetting that T = F, because the thread has no mass. The tension at the rod equals the friction at the ball.
- Using the climb l = ½w₂t² for the ball alone. The relative acceleration w₁ − w₂ is the one that matters.
- Putting the sign the wrong way. If M were less than m, the formula would give a negative F, which means the ball could not rise on the rod.
- Assuming the ball always falls. If F is above mg the ball goes up while the rod still falls.