Where does the ball hit the floor?

Question

A 110 g ball on a 60 cm long string is swung in a vertical circle about a point 200 cm above the floor. The tension in the string when the ball is at the very bottom of the circle is 3.5 N. A very sharp knife is suddenly inserted, as shown in Figure P7.51, to cut the string directly below the point of support. Where does the ball hit the floor? (Measure from the point where the string was cut and consider right to be the positive direction.)

Answer 1

Gee if I only could see that Figure P7.51.

So let me do the next thing and guess.

I guess the string was cut as it was vertical and the ball was facing the floor.

It has a tangential or horizontal velocity V
V=sqrt(a R)
a=(F-W)/m = (F - mg)/m=F/m -g
R=.6m

For a free fall
t=sqrt(2h/g)
S=Vt=
S=sqrt(a R) sqrt(2h/g)
S=sqrt(a R 2h/g)
S=sqrt(2[F/m -g] R h /g)
S=sqrt(2 [ 3.5N/ .110 - 9.81] 0.6 x 0.200 /9.81)
S=0.734 m
or 73.4 cm

Answer 2

First, you must draw a force diagram for the ball and determine the net (centripetal) force. In this case, the net force is calculated by Fnet = Tstring - mg.

Then, you take Fnet and plug it into the formula from Newton's 2nd law in order to determine the tangential acceleration of the ball. This formula is: Fnet = (m)(v^2)/r. This gives you the velocity (v) in the x-direction.

The next step is a 2-dimensional kinematics problem, where the motion in the y-direction is freefall and the motion in the x-direction must be determined using the kinematics equations. Use the motion in the y-direction to solve for the unknown time at which the ball hits the ground using the formula Yf = Yi + Vi(t) + 1/2Ay(t^2). Then plug the time value into the kinematics formula Xf = Xi + Vi(t) + 1/2ax(t^2) in order to find the final x position.