I had this discussion with Ron before, but he didn't believe me.
(I corrected the units with the help of a fellow poster)
Do balls with different RGs have different initial ball speeds?
I read on a several web sites that the RG of a round object is equal to the square root of the Moment of inertia divided by weight.
If I understand this correctly then RG = (IM/BallWeight)^0.5. where the term ^0.5 means the square root for you non-engineering folk.
Thus: IM^0.5=RG*BallWeight^0.5 or IM = RG^2 * Ball Weight
In typical US units, IM or Moment of Inertia has the units lbm-ft^2 (apparently these are in lbm-in^2.)
It is strange to think that Rg has units of feet (again in inches not feet), If that is the case, then I would think that the manufacturers would have to specify the units of RG, because the number would be different for US units and SI units.
An Rg of 2.6 in would be equal to 0.06604 meters. Since ball manufacturers in the US give us ball weights in pounds, then I will assume for this exercise that RG is measured in US units and thus lbm-in^2
So lets say you have a 15 lb ball with an RG of 2.6, thus the IM would = 2.6^2 * 15 or 101.4 lbm-in^2 . The equivalent Moment of Inertia for a ball with an RG of 2.45 would be 90.0375.
For a bowler who puts 300 RPM on a typical throw, let us figure out the Torque that he is applying.
Since Torque = IM * alpha then alpha = Torque/IM. Unfortunately alpha is radial acceleration, and not radial speed or RPMs.
So what is the difference between acceleration and speed? Well lets just assume that in our experiment that the bowler applies a torque, and accelerates the ball from 0 RPM to 300 RPM (as it leaves his fingers) in one tenth of a second. Thus alpha = ( Initial Speed - Final Speed) / time or (0-300)/0.1 = 3000 rpm/second or 50 revs/second^2 ... x 2 pi = 314 rad/s^2.
Thus Torque for the 2.6 Rg ball would be ... (101.4 lbm-in^2) * (314 rad/s^2) = 31,730.08 lbm-in^2- rad/ sec^2
Since US units are really screwy, let me tell you that typically torque is force through a distance or lbf-ft. 1 lbf = 32.2 lbm-ft/sec^2 So torque is 1 lbf-ft = 32.2 lbm-ft^2/sec^2
After all conversions, the torque would be 6.87 lbf-ft
Since the torque that we apply on the ball is the same, for a good bowler with consistent release, regardless of the ball moment of inertia, let us see if we get a different speed for the lower RG ball.
With T = 6.87 and IM = 90.0375 then alpha is 56.308 revs/s^2 = 353.79 rad/s^2.
With the same 0.1 time of force application, then the final velocity leaving the bowlers hand would be ... 337 RPM.
So an RG diff of two balls with 2.45 to 2.6 would equate to revs of 337 to 300. Thus a 6.1% RG diff is equivalent to 12.3% increase in RPMs
Now we both know that a bowling ball is a little more complicated that this. The RG is not fixed, and depending on where you put the pin, and how you drill an extra hole, you do vary the moment of inertia. But in general, I feel fairly confident about these calculations.
One thing I wasn't sure of at first was using 0.1 seconds as the period of time to accelerate the ball, but I feel that this is ok. If I had chose 0.2 or 0.5 or any other number it really wouldn't matter in the calculation, as long as a I used the same number for both balls. The ball is accelerated only while the force is being applied.
I guess I could be talked into thinking that the period of time that it takes to accelerate a lower RG ball would be quicker, and thus would leave the persons hand sooner, but I don't think that this is the case either.
Small reference below:
Q: Please could you define for me the Radius of Gyration in the context of a Flywheel? Thankyou.
A:
The radius of gyration is defined as (I/M)^1/2, or the square root of the moment of inertia divided by mass. The moment of inertia is an important thing to know when solving problems that have to do with how things rotate. The equation for moment of inertia is different depending on the shape of the object, but for a flywheel (basically a solid disk), it's I=1/2MR^2, where M is the mass and R is the radius of the flywheel. Putting this into the equation from before, we find that the radius of gyration for a flywheel is R*(1/2)^1/2, or 0.707*R.
The units of moment of inertia are mass*distance^2, so if you divide by mass and take the square root, you get something with units of just distance (a.k.a. the radius of gyration). Conceptually, the radius of gyration is the distance that, if the entire mass of the object were all packed together at only that radius, would give you the same moment of inertia. That is, if you were to take the entire mass of the disk-shaped flywheel with some radius and pack it into a narrow donut whose radius is the flywheel's radius of gyration, they'd both have the same moment of inertia. With the same moment of inertia, they will behave very similarly when you spin them.
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dR3w
"I did nothing. I did absolutely nothing, and it was everything I thought it could be. "
Edited on 4/16/2008 10:53 AM
