Ball on a String?

Question

A ball held by a string is coasting around in a large horizontal circle. The string is then pulled in so the ball coasts in a smaller circle. When it is coasting in the smaller circle its speed is:

a) greater
b) less
c) unchanged

Yes - I know it's not a difficult question but it allows for a great discussion of the mechanics involved and relates to the last question nicely.

The answer is a, greater. When the ball is moving in a circular path of constant radius the pull of the string does not speed it up -- it coasts at constant speed. But when the ball is pulled into a smaller circle, the force of the string speeds it up. Why? Because in the first case where the radius remains constant the string is always pulling at right angles to the motion of the ball -- in other words the pull is always sideways and serves only to keep the balls direction of motion changing -- into a circular path instad of a straight line (which would be the case if the string broke).

But when the ball is being pulled in from a large circle to a smaller circle the force of the string is not exactly sideways to the ball's motion. We can sketch it or visualize that the force of the string ha a component of force in the direction the ball is moving. Component 1 acts to increase the speed, and Component 2 simply changes the ball's direction of motion.

There is an ingenious way ...

to determine this increase in speed. It depends on the idea of angular momentum, which for a bod of mass m moving at speed v at a radius r is the product mvr. Just as a force is required to change the linear momentum of a body -- if no torque acts on a rotating body, its angular momentum cannont change. Does the string exert a torque on the ball? It does only if the force acts through a lever arm (recall my prior question TORQUE). Is there a lever arm? No, the line of action of the force passes straight through the pivot point. There is no torque, and therefore no change in angular momentum. The ball has the same angular momentum mvr when it arrives at the small circle that it had on the large circle. Since the product mvr doesn't change, any decrease in radius is exaclty compensated by an increase in speed. Say if r decreases by half, then v increases by two; if r is pulled in to a third its initial value, the speed triples. This holds true in the absence of a torque, which would...

otherwise change the angular momentum.

We can represent this ideapicorially, just as we did for torque. Recall that torque is force x lever arm, and can be pictured as twice the area of the triangle formed by the force vector and the lever arm. So too, angular momentum = mvr = (linear momentum mv) x (lever arm r) can similarly be represented by twice the area of the triangle formed by the linear-momentum vector and the lever arm. If no torque acts on any rotating system, its angular momentum will not change. In the case of our circling ball, the angular momentum is the sameon the outer circle as it is on the inner circle. That means the area of the triangle formed by the linear momentum and the pivot point doesn't change. Like if the radius of the smaller circle is one third the radius of the larger circle, then the linear momentum (mv) of the ball on the smaller circle must be three times larger than when on the large circle. That means it goes three times as fast.

Note that in this way the area of the angular momentum triangle comes out to be the same for both circles.

Answer 1

Greater.
When you pull it inwards you are applying a force that is always perpendicular to its velocity, so you don't change the velocity that way directly. However, you are doing work on the system since it takes force to pull in the string.
We know that angular momentum is conserved because you can't apply a torque about the center by a floppy string. Angular momentum H is defined as the moment of inertia I * the angular velocity w; H = Iw. I = mass * radius^2 = mr^2. Multiplying r by 0.5 multiplies r^2 by 0.25 and since m is constant I is multiplied by 0.25. For H to stay constant w must increase by 4. Since velocity v = wr, v increases by 2. So through this tortuous path, work pulling in the string becomes converted to kinetic energy in the form of a velocity increase.

Answer 2

Greater becoz
When u pull it inwards you are applying a force that is alwys prpendiclr to its velocity, so you don't change the velocity that way directly. However, you are doing work on the system since it takes force to pull in the string.

Answer 3

hi you could always try a DIY shop or you could always try a shop called woolworths they may have a ball of string. you can always try on line as well. there is a shop called hobby craft but i am not sure if there is one in london.

Answer 4

greater