Question

A small 0.14 kg metal ball is tied to a very light (essentially massless) string that is 0.9 m long. The string is attached to the ceiling so as to form a pendulum. The pendulum is set into motion by releasing it from rest at an angle of 55 ∘ with the vertical.(a) What is the speed of the ball when it reaches the bottom of the arc?(b) What is the centripetal acceleration of the ball at this point?(c) What is the tension in the string at this point?

Answer 1

Answer:

a) v=2.743m/s

b) $a_c = 8.363m/s^2$

c) T=2.543N

Explanation:

First, calculate the height of the ball at the starting point:

$y' = 0.9cos(55)$

$y' = 0.516$

At this point, just in the moment the ball is released, all the energy of the system is potencial gravitational energy. When it is at the bottom all the potencial energy is transformed into kinetic energy:

$E_p=E_k\\mgh=\frac{mv^2}{2}$

Solving for v:

$v=\sqrt{2gh}$

if h is the height loss: (l-y')

v=2.743m/s

The centripetal acceleration is the acceleration caused by the tension force exercised by the string, and is pointing outside of the trayectory path (at the lowest point, directly dawn):

$a_c=\frac{v^2}{r}$

$a_c = 8.363m/s^2$

To calculate tension, just make the free body diagram of forces in the ball, noticing the existence of the centripetal acceleration:

$\sum{F_y}=ma_c=T-W\\T=ma_c+W\\T=m(a_c+g)\\T=0.14(8.363+9.8)\\T=2.543N$