4 ms·
> For a twice larger "spin resistor", then, the equilibrium torque would be twice as large, and intuitively, the chain speed would be half as fast. For the twi
by gugagore 4y ago
> For a twice larger "spin resistor", then, the equilibrium torque would be twice as large, and intuitively, the chain speed would be half as fast.
For the twice-larger resistor, the torque is the same, and the speed is half as fast. Can you explain why you think the torque is twice as large?
To move the 2x resistor at the same speed as the 1x resistor, you need twice the torque.
- kayson 4y agoMaybe my misunderstanding is how the speed would be independent of torque. My thought was that the 2x resistor has 2x the coefficient of kinetic friction, which means 2x force and 2x frictional torque. In order for the system to be at equilibrium, meaning constant speed, the battery torque must match the friction torque.