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Hi,

For a project, I am making a badminton shuttlecock launching machine. Similar to other pitching machines like tennis, the shuttle in this case, travels through two spinning disk rotating in opposite direction, and gets launched. I am having trouble picking the type of motor that will be required to throw the shuttle at 54 m/s.

The two spinning disks are 0.080 meters in Diameter. According to my calculations to achieve 54 m/s linear velocity, It will need to rotate at 13000 RPM. but I cant seem to find the power or torque required for the motor. Please guide me to find them.

If needed, the mass of each disk is 0.1 kg

the mass of shuttle is 0.005 kg

Pranav, Thanks for your question. I will have a full response to your question shortly. My analysis indicates that you should need an electric motor with only a modest torque and power capability. A web site for appropriate electric motors is given here:

http://www.google.com/#q=electric+motors&hl=en&source=univ&tbm=shop&tbo=u&sa=X&e.

A motor with less than 1 horsepower should be able to spin up your disks to the desired RPM ( ~ 13000 RPM with appropriate gearing) and to recover their speed quickly after each shuttle is launched. Motors with this capability appear to be on the order of $100 or less.

Hope this helps.

Randy

3/25/13 Here are some further results for your problem.

The key to this is to figure out how the linear momentum needed for the birdie is extracted from the angular momentum of the spinning disks. As the birdie slips between the disks, its increase in momentum can be treated as an impulse (torque integrated over a short time) and equated to the angular momentum of a particle on the radius, R, of the disks. Let vb be the velocity desired for the birdie, then the change in angular momentum is

La = (vb/R)mR where m = mass of birdie and vb actually represents the change in velocity from 0 to vb .

The disks have to supply this momentum by suddenly slowing down, losing

I(vi-vb)/R where I = moment of inertia of the disks = 8 x 10-^5 kgm^2 and vi = initial tangential velocity before the birdie enters. Equatiing the momenta and solving for vi gives

vi = vb( (mR^2/I + 1) = (2.25)(54 m/s) = 121.5 m/s. The RPM would need to be 2.25 times the 13000 needed for the birdie speed or about 1300 rad/s.

Now in a frictionless world, the disks could be spun up to this speed, the birdie would then squirt through and reduce the speed and everything would then stay the same. However, the disks have to be spun up to receive the next birdie, and this is where the torque comes in. If we want the spin up to occur in 1 second, then

τ = ∆L/∆t = mRvb/(1 sec) = (0.005kg)(0.004m)(54m/s) = 1.08x10^-3 kg(m/s)^2

in terms of power P = τω = (1.08x10^-3)(1300) = 1.3 Nm/s = 1.3 watts ≈ (1/646) horsepower ≈ 1.3x10-3 HP, which isn't very big. So what you need is a motor than can

a) spin up the disks to the desired RPM

b) produce a torque at around 20000 RPM that can quickly ramp up the speed after launching a birdie.

The motors listed in the website shown above have ~ 1 HP and RPM ~ few 1000. This should be enough even given the need for gearing needed to achieve the RPM (a factor of 2 in RPM requires a factor of 2 increase in available torque).

Let me know if this works for you.

Randy

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