Answer:
L = 4.574 ft
Explanation:
Given:
- The weight of the block W = 2 lb
- The initial velocity of the block v_i = 0
- The stiffness of the spring k = 2 lb/ft
- The radius of the cylindrical surface r = 2 ft
Find:
Determine its unstretched length so that it does not allow the block to leave the surface until θ= 60°.
Solution:
- Compute the velocity of the block at θ= 60°. Use Newton's second equation of motion in direction normal to the surface.
F_n = m*a_n
Where, a_n is the centripetal acceleration or normal component of acceleration as follows:
a_n = v^2_2 / r
- Substitute:
F_n = m*v^2_2 / r
Where, F_n normal force acting on block by the surface is:
F_n = W*cos(θ)
- Substitute:
W*cos(θ) = m*v^2_2 / r
v_2 = sqrt ( r*g*cos(θ) )
- Plug in the values:
v^2_2 = 2*32.2*cos(60)
v^2_2 = 32.2 (ft/s)^2
- Apply the conservation of energy between points A and B where θ= 60° :
T_A + V_A = T_B + V_B
Where,
T_A : Kinetic energy of the block at inital position = 0
V_A: potential energy of the block inital position
V_A = 0.5*k*x_A^2
x_A = 2*pi - L ..... ( L is the original length )
V_A = 0.5*2*(2*pi - L)^2 =(2*pi - L)^2
T_B = 0.5*W/g*v_2^2 = 0.5*2 / 32.2 *32.2 = 1
V_B = 0.5*k*x_B^2 + W*2*cos(60)
x_B = 2*0.75*pi - L ..... ( L is the original length )
V_B = 0.5*2*(1.5*pi - L)^2 + 2*1 = 2 + ( 1.5*pi - L )^2
- Input the respective energies back in to the conservation expression:
0 + (2*pi - L)^2 = 1 + 2 + ( 1.5*pi - L )^2
4pi^2 - 4*pi*L + L^2 = 3 + 2.25*pi^2 - 3*pi*L + L^2
pi*L = 1.75*pi^2 - 3
L = 4.574 ft
