Transformations of Energy in a Pendulum Type Experiment

[Khaled332]

Homework Statement

The string in Fig. 8-35 is L = 120 cm
long, has a ball attached to one end, and is
fixed at its other end. The distance d to the
fixed peg at point P is 75.0 cm. When the
initially stationary ball is released with the
string horizontal as shown, it will swing
along the dashed arc. What is its speed
when it reaches
1.What is the tension of the string at the final position (answer in terms of m, g, L, a)
2. What is the minimum value ofd so the ball will reach the final position (answer in the same terms)

Link: http://www.nevis.columbia.edu/~sciulli/Physics1401/homeworks/HW4.pdf

Homework Equations

Ep=mgL ; Ek=mv^2/2

The Attempt at a Solution

1. Fu=T-mg
Fu=T-9.8a
ma=T-9.8a
T=ma+9.8a ? T=ma+ga

2.Cannot understand
Thanks

 

[anathema]

for posting this problem! Let's break it down and solve it step by step.

1. To find the tension of the string at the final position, we can use the conservation of energy principle. At the initial position, the ball has only gravitational potential energy, which is equal to Ep = mgL. At the final position, the ball has both kinetic and potential energy, which is equal to Ek + Ep = mv^2/2 + mgL. Since energy is conserved, we can set these two equations equal to each other:

Ep = Ek + Ep
mgL = mv^2/2 + mgL
mgL - mgL = mv^2/2
0 = mv^2/2
v^2 = 0
v = 0

This means that the ball will reach the final position with zero speed, since all of its initial potential energy has been converted to potential energy. Therefore, the tension of the string at the final position is simply equal to the weight of the ball, which is T = mg.

2. To find the minimum value of d so that the ball will reach the final position, we can use the conservation of energy principle again. At the initial position, the ball has only gravitational potential energy, which is equal to Ep = mgL. At the final position, the ball has both kinetic and potential energy, which is equal to Ek + Ep = mv^2/2 + mgL. We also know that the ball will reach the final position with zero speed (as we found in part 1), so the final kinetic energy is equal to zero.

Setting these two equations equal to each other and solving for d, we get:

Ep = Ek + Ep
mgL = 0 + mgL
mgL = mgd
d = L

Therefore, the minimum value of d is equal to the length of the string, L. Any value of d greater than or equal to L will result in the ball reaching the final position.