Question

The 2-lb block is released from rest at A and slides down along the smooth cylindrical surface. Of the attached spring has a stiffness k = 2 lb/ft.
Determine its unstretched length so that it does not allow the block to leave the surface until θ= 60°.

Answer 1

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