A mass hangs on the end of a massless rope. The pendulum is held horizontal and

Question

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 = 2.7 m/s and the tension in the rope is T = 20.2 N.   

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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 = 2.7 m/s and the tension in the rope is T = 20.2 N. How long is the rope? What is the mass? If the maximum mass that can be used before the rope breaks is mmax = 1.59 kg, what is the maximum tension the rope can withstand? 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)?

Explanation / Answer

1. v^2=2gL
L= v^2 / 2g

=2.7^2/2g

=0.371m

2. M= LT/(v^2 + Lg)

=(0.371*20.2)/(7.29+3.63)

=0.6862 Kg

3. T = Mv^2 / L + Mg

T={(1.59*2.7^2)/0.371}+(1.59*9.81)

T=46.83N

4. V = (2gH)^1/2

=(2*9.8*0.296)

=2.41

5. T = Mv^2 / L + Mg

= (0.6862*2.41^2/0.074) + 6.72

=60.72N

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