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 = 2.1 m/s and the tension in the rope is T = 18.4 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 = 1.49 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

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 = 2.1 m/s and the tension in the rope is T = 18.4 N. How long is the rope? What is the mass? If the maximum mass that can be used before the rope breaks is m max = 1.49 kg, what is the maximum tension the rope can withstand? (Assuming that the mass is still released from the horizontal.)

Explanation / Answer

1)energy conservation:

mgL = 0.5mv^2

=> L = v^2/(2g) = 0.225m

2)T - mg = mv^2/R

=> m = T/(g+v^2/R) = 0.626 kg

3)Tmax = m(g+v^2/R) = 14.9(9.8 + 2.1^2/0.225) = 438.06 N

4)0.5mv^2 = 0.5mv'^2 + mg(L/5)

=> 2.1^2 = v^2 + 0.4*9.8*0.225

=> v = 1.878 m/s

5) T = mv^2/(L/5) = 0.626*(1.878^2)/(0.225/5) = 49.06 N

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