Problem 3/73 Page 148, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition
Given: A ball of mass m moving in a horizontal circle at speed v. The ball is supported by a string of length L attached to a point a distance h above the center of the circle.
Find: h and the tension T in the string.
1. Mechanical System: Ball at instant shown.
2. Free Body Diagram: Not included. Would show ball, weight of ball acting vertically downward, tension of string acting along the line from the ball to the attachment point. Coordinate axes would be shown adjacent to the free body diagram. The n direction would be shown in the horizontal plane of the circle directed toward the center of the circle. The t direction would be shown in the horizontal plane tangent to the circle. The z(k) direction would be shown vertically upward.
3. Equations:
S F = { T h/L - m g } k +
{ T (L2-h2)1/2/L } n
m a = m ( at t + an n ) = m ( dv/dt t +
v2/r n ) = m ( dv/dt t
+ v2/(L2-h2)1/2 n
)
4. Solve:
Considering the k and n directions:
T h/L - m g = 0
T (L2-h2)1/2/L = m v2/(L2-h2)1/2
Solving the top equation for T in terms of h:
T = m g L / h
Dividing the second equation by the radical and using the result for
T:
m g / h = m v2 / (L2 - h2)
Dividing both sides by m g and multiplying both sides by h(L2-h2)
and grouping terms to reveal a quadratic equation in h:
h2 + (v2/g) h - L2 = 0
Solving for h (requiring h to be positive):
h = -1/2 (v2/g) + { 1/4 (v2/g)2
+ L2 }1/2
Plugging back into the equation for the tension T:
T = m g / [ -1/2 (v2/g L) + { 1/4 (v2/gL)2
+ 1 }1/2 ]
Note that the above appear to be different from those given in the
book. To see the equivalence note that the variable w
introduced by the book is v/r. However r is equal to (L2-h2)1/2.
Thus v2/(L2-h2) = w2.
Using this relationship the equivalence in the expressions can be demonstrated.