Cam follower stall at the nose
A student is building a mechanical display for a science fair. A 3 V geared motor runs at 60 rpm and has a stated stall torque of 0.08 N·m. The motor drives a pear-shaped cam that lifts a follower and a 120 g cardboard lever by 25 mm; a return spring pulls the follower down. At first the mechanism works for about 20 cycles, then the follower sticks at the highest point and the motor stalls. The cam is laser-cut from 3 mm acrylic, the follower is a 6 mm steel rod with a flat face, and the spring is a 0.5 N/mm compression spring compressed by 12 mm at the top of the lift. The cam profile has a 15 mm base circle and a 25 mm lift measured from the base circle. The student wants to keep the same motor and cam lift.
1. Analyse why the follower sticks at the top of the lift and the motor stalls. Use the forces and geometry, not just 'friction'.
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At the top of the lift the spring is compressed 12 mm, so its force is 0.5 N/mm × 12 mm = 6 N. The 120 g lever adds about 1.2 N, giving a total follower load of roughly 7.2 N. The cam contact is near the nose, about 15 mm base radius + 25 mm lift = 40 mm from the cam centre. The moment required to hold the follower is therefore about 7.2 N × 40 mm = 288 N·mm = 0.288 N·m. The motor's stall torque is only 0.08 N·m, so at the nose the motor cannot supply enough torque and stalls. Earlier in the cycle the spring compression is smaller and the contact radius is closer to the base circle, so the required torque is lower; that is why it ran for about 20 cycles before sticking at the highest point. The flat-faced follower also creates side thrust and friction, and the acrylic cam may have worn, increasing friction. The stall is a torque-demand problem at the nose, not simply 'the motor is too weak everywhere'.
2. Recommend two changes to the mechanism that keep the 3 V motor and the 25 mm lift. Explain how each change reduces the required torque.
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Two changes that keep the motor and 25 mm lift:
- Reduce the spring force or preload. Use a spring of about 0.1–0.15 N/mm, or reduce preload, so the return force at the top is only about 1.2–1.8 N plus a margin. The spring only needs to overcome friction and lever weight, not press hard against the cam. This cuts the nose torque roughly in proportion, from about 0.29 N·m to well under 0.08 N·m.
- Use a roller follower and polish or lubricate the cam. A roller follower replaces sliding friction with rolling contact, reducing side thrust and wear. A 6 mm steel flat follower on acrylic has high friction; a small ball-bearing roller or a polished steel cam reduces the torque lost to friction. If a roller is not available, a rounded follower face and a PTFE-loaded acetal cam can help.
A third option is to counterbalance the lever so the spring only returns it, but that adds mass. Increasing motor voltage is not a mechanism fix. The student should test the torque at the nose with a spring scale or a small torque wrench before cutting a new cam.
3. The student suggests increasing the motor supply from 3 V to 4.5 V to get more torque. Evaluate this suggestion, including one risk.
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Increasing the supply from 3 V to 4.5 V will increase the motor's speed and stall torque roughly in proportion, so it may appear to solve the stall. But the root cause is that the required torque at the nose is about 0.29 N·m versus a 0.08 N·m motor; a 50% voltage increase gives only about 0.12 N·m, still short. The motor will run closer to stall, draw a high current, heat up and may burn out or wear brushes quickly. It may also overspeed the cam and increase impact and friction. The safer engineering solution is to reduce the load (spring force, friction) rather than overdrive the motor. If the student does try 4.5 V, they should check the motor's rated voltage, measure current, add a fuse, and test for overheating. A flywheel could help carry the mechanism through the nose, but it does not reduce the static torque demand and could make the mechanism harder to stop.