A mass hangs on the end of a massless rope. The pendulum is held horizontal and released from rest. When the mass reaches the bottom of its path it is moving at a speed v = 2.4 m/s and the tension in the rope is T = 14.9 N.
1) How long is the rope?
2) What is the mass?
3) If the maximum mass that can be used before the rope breaks is mmax = 1.2 kg, what is the maximum tension the rope can withstand? (Assuming that the mass is still released from the horizontal.)
4) Now a peg is placed 4/5 of the way down the penduum.

L = 0.294 m

m = 0.507 kg

T = 35.3 N

T = 8.77 N

Explanation:

Given:

- The mass is released from rest vi = 0 m/s

- The final velocity of mass at bottom vf = 2.4 m/s

- The Tension in the rope T = 14.9 N

Find:

1) How long is the rope L?

2) What is the mass m?

3) If the maximum mass that can be used before the rope breaks is mmax = 1.2 kg, what is the maximum tension the rope can withstand? (Assuming that the mass is still released from the horizontal.)

4) Now a peg is placed 4/5 of the way down the pendulum.

Solution:

- First apply the principle of conservation of energy at initial and final states as follows:

                              K.E_1 + P.E_1 = K.E_2

                              0.5*m*vi^2 + m*g*L =  0.5*m*vf^2

                              0 + m*g*L = 0.5*m*vf^2

                              L = 0.5*vf^2 / g

                              L = 0.5*2.4^2 / 9.81

                              L = 0.294 m

- Apply Newton's second Law of motion in bottom most along centripetal direction, with centripetal acceleration a_c:

                              T - m*g = m*a_c

                              a_c = vf^2 / L

                              T = m*vf^2 / L + m*g

                              m = T / (vf^2 / L + g )

                              m = (14.9)/ [ 2.4^2 / 0.294 + 9.81 ]

                              m = 0.507 kg

- For the case the maximum mass m_max required for the mass to break-off the string. Then we have the following Newton's law of motion:

                              T = m_max( vf^2 / L + g )

                              T = 1.2*(2.4^2 / 0.294 + 9.81 )

                              T = 35.3 N    

- The peg is placed at 4/5 of the path way(arc length) the angle swept is θ.

                              s = L*θ

                              4*pi*L/2*5 = L*θ

                              θ = 2*pi/5

- The vertical height h at position θ = 2*pi/5 is: Trigonometric relation

                              h = L( 1 - sin(2*pi/5))

                              h = 0.04894*L

- Apply conservation of energy and find the velocity vp when mass m hits the peg.

                             K.E_1 + P.E_1 = K.E_2 + P.E_2

                             0.5*m*vi^2 + m*g*L =  0.5*m*vp^2 + m*g*h

                             0 + m*g*L = 0.5*m*vp^2 + m*g*0.04894*L

                             vp^2 = 2*g*L*(0.95105) = sqrt (2*9.81*0.294*0.95105)

                             vp = 2.3422 m/s

- Apply the Newton's second Law of motion when mass m hits the peg.

                             T - m*g*sin(2*pi/5) = m*vp^2 / L

                             T =  m*vp^2 / L + m*g*sin(2*pi/5)

                             T = 0.507*[2.3422 / 0.294 + 9.81*sin(2*pi/5)]

                             T = 8.77 N