# Creation of Rotational K.E.?

by greswd
Tags: creation, rotational
 P: 147 I was thinking about using a power drill to spin up a gyro. Let's say this gyro has two segments, a wide center and narrow ends. And its a perfectly rigid body. When the power drill is placed at the wide part, more torque is applied to the gyro and the gyro's angular acceleration is greater. When the power drill is placed at the narrow part, less torque is applied to the gyro and the gyro's angular acceleration is much less. In both cases the torque the gyro applies on the power drill should be the same. Then we have a dilemma, because in both cases the drill loses the same amount of energy, while in the first case the gyro has more rotational energy than in the second case. i must have overlooked something, but I can't seem to think of it now.
 HW Helper P: 7,118 Assuming the gyro continues remains attached at the same point of the power drill bit, then when the drill is in the narrow part, the gyro is rotating about a different axis, and there is more angular inertia. Assuming the process is lossless, then the increase in angular energy will equal the work done by the power drill. For a better comparason, compare the cases where the power drill is connect to the center of the axis of the gyro, versus the case where the power drill uses a massless "tire" to drive the gyro from the edge, with the gyro spinning about it's central axis.
 Sci Advisor P: 2,470 Energy transfer per unit time (power) is equal to torque times angular velocity. If one spins up faster than the other, there is more power used as well. You shouldn't assume that equal force/torque gives you equal power. If I push against the wall, no matter how hard, if the wall doesn't move, no work is being done and no energy is transferred.
 Engineering Sci Advisor HW Helper Thanks P: 7,150 Creation of Rotational K.E.? Thinking about the power drill may be confusing, because simple "common sense" assumptions about how the drill converts electrical energy into mechanical work and/or heat, and how much electrical energy the drill actually uses (which will depend on its RPM) are probably wrong. Think about a simpler method of spinning up the gyro, for example winding a string aroud the shaft or the outside of the wheel, and tying a weight to the end of the string. If the weight is pulled down by the force of gravity and the string unwinds, the amount of work done only depends on the distance the weight falls. The rate of acceleration, and the time it takes to unwind the string, will depend where you wind the string (round a large or small radius), but for the same length of string, the final angular velocity will be the same. Even better - don't just think about it, do the experiment yourself and find out what happens!
HW Helper
P: 7,118
 Quote by greswd I was thinking about using a power drill to spin up a gyro. Let's say this gyro has two segments, a wide center and narrow ends. And its a perfectly rigid body. When the power drill is placed at the wide part, more torque is applied to the gyro and the gyro's angular acceleration is greater. When the power drill is placed at the narrow part, less torque is applied to the gyro and the gyro's angular acceleration is much less.
After looking at the other responses, it's not clear to me what you are asking here. One case is that the power drill's axis is perpendicular to the gyro's axis, and you effectively have a "tire" at the end of the power drill to spin the gyro? Another case is that the gyro's axis is common to the gryo's axis and directly driving the gyro. Another case is that the drill's axis is parallel to the gryo's axis, and that the gryo drill bit is acting as a "tire" driving the gryo from the outer edge of the gyro.

So for this problem, how is the power drill driving the gryo from the thick part and from the narrow part?
P: 147
 Quote by K^2 Energy transfer per unit time (power) is equal to torque times angular velocity. If one spins up faster than the other, there is more power used as well. You shouldn't assume that equal force/torque gives you equal power. If I push against the wall, no matter how hard, if the wall doesn't move, no work is being done and no energy is transferred.
Oh yeah thanks. If the power drill is applied to the wider part, it takes less time to reach a certain angular velocity, but the same amount of work is done.
P: 2,470
 Quote by greswd but the same amount of work is done.
Not if the torque was constant, as you said. Though, torque of a drill depends on RPM, so it's not an assumption you should make.

Keep in mind that total amount of work done must match total rotational K.E. at the end, and that will depend on moment of inertia. If more torque is required to impart certain angular acceleration, then the moment of inertia is higher. More work will be done on that object to get it up to target speed. Under constant torque, or even fixed torque/RPM curve of a drill, this will mean more time.
P: 147
 Quote by K^2 Not if the torque was constant, as you said. Though, torque of a drill depends on RPM, so it's not an assumption you should make. Keep in mind that total amount of work done must match total rotational K.E. at the end, and that will depend on moment of inertia. If more torque is required to impart certain angular acceleration, then the moment of inertia is higher. More work will be done on that object to get it up to target speed. Under constant torque, or even fixed torque/RPM curve of a drill, this will mean more time.
No I said constant torque on the drill, but different torques on the gyro.

The thing that is constant for all is the force.
 P: 3,143 About the only known constant (in the problem) is the moment of inertia of the giro. If we wanted to accelerate it at the same rate using different "gear" ratios we know that the torque required would be different. But how is it actually accelerating? Is it a cordless DC drill or an AC drill? I think most AC drills use a universal brushed motor.. http://www.johnsonelectric.com/en/re...rs-theory.html Looking at the chart.. When stationary the operating point is at the right hand side of the chart (eg min speed). As speed increases the operating point moves towards the middle where you could say the output power curve is relatively flat and independant of torque and rpm. So perhaps safe to model it as is a constant power device when loaded? The energy stored in a flywheel is E = 0.5*I*ω2 Power = Energy/time so Power = 0.5*I*ω2/t Acceleration is ω/t (eg not ω2/t) so it's not going to accelerate at constant rate.

 Related Discussions Introductory Physics Homework 5 Introductory Physics Homework 3 General Physics 3 Advanced Physics Homework 0 General Physics 14