Dear Don,
You can't be too careful with getting physics right! The above calculations really assume the ball just drops in free fall, but in actuality it rolls. When the rolling (versus dropping) is observed, the distinction between rotational speed and translational speed of the ball's center is critical. At the bottom of the ramp, the translational speed is what you are asking about.
The rolling of a ball down an inclined plane does not accelerate the translational speed the same way that a block (for example) would slide without friction down a ramp. The rolling reduces the friction between the ball and the ramp so that friction does not play much of a role in affecting the acceleration (and this is why Gallileo used a ball and ramp to isolate the effect of gravity from that of friction). But if you compare a frictionless block sliding down a ramp and a ball rolling down the same ramp, the translational acceleration of the block is faster than that of the ball, and the resulting velocity at the bottom of the ramp is accordingly faster for the block than the ball.
This
SparkNotes physics website explains the distinctions and gives the correct formula. The upshot is that the velocity at the bottom of the ramp of a rolling DISK is given by:
v = SQR(g
h 4/3)
where g = acceleration force of gravity, 32.2 feet per second per second, or 384 inches per second per second;
where h = height of ball at release (10.6 inches)
The explanation is given:
"At the top of the incline, the disk has no kinetic energy, and a gravitational potential energy of mgh. At the bottom of the incline, all this gravitational potential energy has been converted into kinetic energy. However, in rolling down the hill, only some of this potential energy becomes translational kinetic energy, and the rest becomes rotational kinetic energy. Translational kinetic energy is given by 1 /2 mv2 and rotational kinetic energy is given by 1 /2 I2. We can express in terms of v and R with the equation = v/R, and in the question we were told that I = 1/2 mR2. We now have all the information we need to solve for v:
My calculator gives the answer as 76.6 inches per second at the bottom of the ramp, and in feet per second that is 6.1, for a rolling DISK.
The MOMENT OF INERTIA (MOI) of a disk or rolling cylinder or wheel is I = 1/2*M*R^2
where R = radius.
But in the case of a BALL or SOLID SPHERE, the MOI is I = 2/5*M*R^2. See Eric Weinstein's Wikpedia for the
MOI of a Solid Sphere versus the
MOI for variosly shaped objects.
Plugging in the sphere MOI in the basic equation,
1/2 * Mv^2 + 1/2 * Iw^2 = Mgh
where w [omega] = v/R; where I = 2/5
M R^2; were R golf ball = 0.84 inches (but which can be ignored since the term cancels out); where g = acceleration of gravity (384 in/sec^2); h = height of release (10.6 in) and M = mass of ball (which can be ignored, since the term cancels out), the formula reduces to:
1/2 * Mv^2 + 1/2 * [2/5
M R^2] * [v/R]^2 = Mgh
[1/2 * Mv^2] + [2/10
M R^2 *v^2 / R^2] = Mgh
1/2 * Mv^2 + 2/10 * Mv^2 = Mgh
5/10 * Mv^2 + 2/10 * Mv^2 = Mgh
7/10 * Mv^2 = Mgh
v^2 = Mgh / M * 10/7
v^2 = gh * 10/7
and finally:
v = SQR(gh * 10/7)
Plugging in the numbers:
v = SQR(384 in/sec^2 * 10.6 in * 10/7)
Thus the velocity of the ball at the bottom f the ramp (v) released from a height of 10.6 inches above the ground is 76.26 inches / second. In terms of feet/second, this is 6.35 ft/sec.
Any sliding or rolling friction will reduce this slightly, as will twisting the ramp out of vertical or the ball's bouncing at the bottom of the ramp. This also assumes an uneventful release that does not add or subtract force in the initial start of the ball's motion down the ramp.
Sorry for the confusion.
Cheers!
Geoff Mangum
Putting Theorist and Instructor
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