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

a). v=4.77
`(m)/(s)`

b). β=80.61

c). E=88.5J

Step-by-step explanation:

a).

`m1=2.7 kg\\L=4m\\v_(1)=8.1 (m)/(s)\\\beta=63`

`(m*v_(1) ^(2) )/(2)+m*g*y_(1)=(m*v_(2)^(2) )/(2)+m*g*y_(2)\\(m*v_(1) ^(2) )/(2)=(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)=√(64-42.8)\\v_(2)=√(21.19)=4.77 (m)/(s)`

b).

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

c).

`E=(m*v_(1) ^(2) )/(2)\\ E=((2.7kg)*(8.1(m)/(s))^(2))/(2)=(177.147(kg*m^(2))/(s^(2)))/(2)\\\\E=88.5 J`