Question

A pendulum consists of a 1.5 kg stone swinging on a 4.3 m string of negligible mass. The stone has a speed of 8.4 m/s when it passes its lowest point. (a) What is the speed when the string is at 64 ˚ to the vertical? (b) What is the greatest angle with the vertical that the string will reach during the stone's motion? (c) If the potential energy of the pendulum-Earth system is taken to be zero at the stone's lowest point, what is the total mechanical energy of the system?

Answer 1

Answer:

(a) v = 1.54m/s, (b) Θ = 80.6° and (c) E = 52.92J

Explanation:

(a) At the lowest point, the total energy is giving by:

$E_{T}_{0} = E_{K}_{0} + E_{P}_{0} = \frac {1}{2} mv_{0}^{2} + mgy_{0}$

where,$E_{K}$: kinetic energy,$E_{P}$: potential energy, m: pendulum's mass, v: speed of the pendulum, g:gravity and y: heigh of the pendulum.

$E_{T}_{0} = \frac {1}{2} mv_{0}^{2} + mg(0) = \frac {1}{2} (1.5)(8.4)^{2} = 52.92 J$ (1)  

When the string is at 64° from the vertical, the total energy is:    

$E_{T}_{f} = \frac {1}{2} mv_{f}^{2} + mgy_{f}$ (2)

At this point, $y_{f}$ is:

$y_{f} = L - Lcos(\Theta)$ (3)

where L: longitude of the pendulum

$y_{f} = 4.3 m (1 -cos(64)) = 2.42 m$

By conservation of energy, we can calculate the speed of the string, at 64 ° to the vertical. Equaling equations (1) and (2):  

$E_{T}_{0} = E_{T}_{f} = \frac {1}{2} mv_{f}^{2} + mgy_{f}$    

$v_{f} = \sqrt \frac{2(E_{T}_{0} - mgy_{f})}{mg}}$  

$v_{f} = \sqrt \frac{2(52.92 - 1.5 \cdot 9.8 \cdot 2.42)}{1.5 \cdot 9.8}} = 1.54 \frac{m}{s}$      

(b) Smilarly, by conservation of energy we can find the greatest angle, assuming that vf = 0 at the greatest angle reached:

$\frac {1}{2} mv_{0}^{2} + mgy_{0} = \frac {1}{2} mv_{f}^{2} + mgy_{f}$

$y_{f} = \frac {E_{0}}{mg} = \frac {52.92}{(1.5)(9.8)} = 3.6 m$    

Using the heigh calculated in equation (3) we can find the angle:

$\Theta = Arccos (\frac {L - y_{f}}{L}) = Arccos (\frac {4.3 - 3.6}{4.3}) = 80.6^\circ$  

(c) The total mechanical energy at the lowest point is giving by:

$E_{T} = E_{K} + E_{P} = \frac {1}{2} mv_{0}^{2} + 0 = \frac {1}{2} (1.5)(8.4)^{2} = 52.92 J$  

Have a nice day!