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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 = 2.3 m/s and the tension in the rope is T = 22.5 N.
How long is the rope?
What is the mass?
If the maximum mass that can be used before the rope breaks is mmax = 1.74 kg, what is the maximum tension the rope can withstand? (Assuming that the mass is still released from the horizontal.)
Now a peg is placed 4/5 of the way down the pendulum’s path so that when the mass falls to its vertical position it hits and wraps around the peg. How fast is the mass moving when it is at the same vertical height as the peg (directly to the right of the peg)?
Return to the original mass. What is the tension in the string at the same vertical height as the peg (directly to the right of the peg)?

Respuesta :

syed514
1) How long is the rope?
0.270m
2)What is the mass?
0.759m

3) If the maximum mass that can be used before the rope breaks is mmax = 1.7 kg, what is the maximum tension the rope can withstand? (Assuming that the mass is still released from the horizontal.)
49.97N

How fast is the mass moving when it is at the same vertical height as the peg (directly to the right of the peg)?
2.06m/s

5)Return to the original mass. What is the tension in the string at the same vertical height as the peg (directly to the right of the peg)?

a = v^2/(1/5)r
After you find acceleration you times it by mass to get the net force of the tension T - 0 = ma --> T = ma


lhmarianateixeira

In this exercise we have to use the pendulum knowledge to determine some characteristics, as we have:

1)0.270m

2)0.507 KG

3) 49.97N

4)2.06m/s

5) 8.77 N

First, we have to organize the information given by the utterance, then:

  • The mass is released from rest [tex]v_i = 0 m/s[/tex]
  • The final velocity of mass at bottom [tex]v_f = 2.3 m/s[/tex]
  • The Tension in the rope [tex]T = 22.5 N[/tex]

1) So calculating the length of the string, we have:

[tex]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*2.3^2 / 9.81\\ L = 0.270 m[/tex]

2) Now to find the value of the mass we have to apply Newton's second law as:

[tex]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)/ [ 2.3^2 / 0.270 + 9.81 ]\\ m = 0.507 kg[/tex]

3) Calculating the system tension we have:

[tex]T = m_max( vf^2 / L + g )\\ T = 1.2*(2.3^2 / 0.270 + 9.81 )\\ T = 49.9N[/tex]

4)  Apply conservation of energy and find the velocity as:

[tex]K.E_1 + P.E_1 = K.E_2 + P.E_2\\ 0.5*m*vi^2 + m*g*L = 0.5*m*v_p^2 + m*g*h\\ 0 + m*g*L = 0.5*m*v_p^2 + m*g*0.04894*L\\ v_p^2 = 2*g*L*(0.95105) = sqrt (2*9.81*0.294*0.95105)\\ v_p = 2.06 m/s[/tex]

5) Apply the Newton's second Law find the tension in the string , we have:

[tex]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[/tex]

See more about pendulum at brainly.com/question/14759840

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