Question

A pendulum consists of a 2.7 kg stone swinging on a 4.0 m string of negligible mass. The stone has a speed of 8.1 m/s when it passes its lowest point. (a) What is the speed when the string is at 63 ˚ 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=4.77 $\frac{m}{s}$

b). β=80.61

c). E=88.5J

Explanation:

a).

$m1=2.7 kg\\L=4m\\v_{1}=8.1 \frac{m}{s}\\\beta=63$

$\frac{m*v_{1} ^{2} }{2}+m*g*y_{1}=\frac{m*v_{2}^{2} }{2}+m*g*y_{2}\\\frac{m*v_{1} ^{2} }{2}=\frac{m*v_{2}^{2} }{2}+m*g*y_{2}\\v_{2}=\sqrt{v_{1}^{2}-2*g*y_{2}}\\y_{2} =L-L*cos(\beta)\\v_{2}=\sqrt{v_{1}^{2}-2*g*(L-L*cos(\beta))}\\v_{2}=\sqrt{8.1^{2}-2*9.8*(4(1-*cos(63))}\\v_{2}=\sqrt{8.1^{2}-42.8}\\v_{2}=\sqrt{64-42.8}\\v_{2}=\sqrt{21.19}=4.77 \frac{m}{s}$

b).

$\frac{m*v_{1}^{2}}{2} +m*y_{1} =\frac{m*v_{2}^{2}}{2} +m*y_{2}\\\frac{m*v_{1}^{2}}{2} +0 =0 +m*y_{2}\\y_{2}=\frac{v_{1}^{2}}{2*g}=\frac{(8.1^\frac{m}{s}){2}}{2*9.8\frac{m}{s^{2} }}=3.34 m\\ 3.34m=L(1-cos(\alpha)\\3.34m=4*(1-cos(\alpha ))\\cos(\alpha )=\frac{4-3.34}{4}=0.16\\(\alpha )= cos^{-1}*(0.163)\\ (\alpha )= 80.61$

c).

$E=\frac{m*v_{1} ^{2} }{2}\\ E=\frac{(2.7kg)*(8.1\frac{m}{s})^{2}}{2}=\frac{177.147\frac{kg*m^{2}}{s^{2}}}{2}\\\\E=88.5 J$