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