Physics Circular motion?

Question

A 60 g ball is tied to the end of a 50-cm-long string and swung in a vertical circle. The center of the circle is 150 cm above the floor. The ball is swung at the minimum speed necessary to make it over the top without the string going slack.
If the string is released at the instant the ball is at the top of the loop, where does the ball hit the ground?

Answer 1

GREAT PROBLEM, WITH FASCINATING ADDITIONAL RESULT(S)! (See below.)

For the string to just not go slack at the top of its swing, the inward centripetal acceleration as the circular motion reaches the very top must be provided entirely by gravity, without ANY tension in the string. At that point, the circular centripetal acceleration, v^2 / r exactly equals the acceleration due to gravity, 'g,' or:

v^2 = g r there, v the speed at the top, r the string's length.

When the ball is released it's moving horizontally, i.e. with NO vertical speed; since vertical and horizontal motions decouple, the time to hit the ground is given by exactly the same expression for free fall from rest:

1/2 g t^2 = 4 r, expressing the height in terms of r. So:

t^2 = 8 r / g.

There's NO acceleration in the horizontal direction; the SQUARE of the distance it will travel horizontally in that time is then:

d^2 = v^2 t^2 = (g r) x (8 r / g) = 8 r^2. ('g' drops out !! )

So d = 2 sqrt (2) r = 100 sqrt (2) cm = sqrt (2) metres.

The ball hits the ground sqrt (2) metres or 1.414... metres to the side of the vertical through the original point of suspension.

ADDITIONAL RESULTS:

1. It's quite amusing that the place where the ball hits the ground is INDEPENDENT OF THE VALUE OF 'g.' You just don't have to quote what the size of 'g' is. If 'g' is much larger, the initial horizontal speed is greater, by a factor sqrt (g); but the time to fall is shorter by the same factor sqrt (g); and the product, which determines how far the ball moves horizontally, remains invariant! That's really a very neat and quite remarkable result.

But that last result itself suggests the following, which can in fact be readily proved:

2. The ENTIRE PATH of the ball, once released, is INDEPENDENT of the value of 'g'!

"Fascinating, Captain." (With one Vulcanian eyebrow archly raised.)

Live long and prosper.

Answer 2

The trick is to ID all the forces on the ball at the instant the string is let go.

At that instant, when the string is released, there is but one force: weight, aka the force of gravity. So as soon as the string is let go the ball begins to fall at g = W/m = 9.81 m/sec^2.

Just prior to that release, there were two forces: weight and centripetal force. But centripetal force disappears as soon as the tension in the string disappears.

So the question becomes, how far in the horizontal direction will the ball fly while it's falling at 9.81 m/sec^2 acceleration.

Discounting drag friction, the horizontal velocity Vh will equal that of the tangential velocity v at the time just before release. To find that, set centripetal force equal to weight C = mv^2/r = mg; so that v^2 = rg and v = sqrt(rg); where r = .5 m, the radius of the circle, and g = 9.81 m/sec^2.

The ball will stay in flight so long as it is dropping; so the time of flight can be found from h = 1/2 gt^2; so that t^2 = 2h/g and t = sqrt(2h/g); where h = 2 m, the height of the ball above ground at release at the top of the circle.

Thus the horizontal distance (S) traveled from the release point is S = vt = sqrt(rg) sqrt(2h/g) = sqrt(2hr) = 2 sqrt(.5) meters. You can do the math.

Physics lesson: the only force acting on a freely moving object is its weight if you discount drag frictional forces. This means that the horizontal component (Vh) of its velocity in flight will remain constant while the vertical component (Vv) will accelerate at g. This follows because of f = ma, which says if a force is zero, there will be no acceleration.

Because, Fh = 0, the horizontal force is zero after release, horizontal acceleration is ah = 0. On the other hand, W = mg = Fv, the vertical force due to gravity is ever present and the ball must accelerate at g = 9.81 m/sec^2 on Earth's surface.