A mass m = 4.6 kg hangs on the end of a massless rope L = 2.03 m long. The pendu

Question

A mass m = 4.6 kg hangs on the end of a massless rope L = 2.03 m long. The pendulum is held horizontal and released from rest.

1)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. As it wraps around the peg and attains its maximum height it ends a distance of 3/5 L below its starting point (or 2/5 L from its lowest point).How fast is the mass moving at the top of its new path (directly above the peg)?

2)Using the original mass of m = 4.6 kg, what is the magnitude of the tension in the string at the top of the new path (directly above the peg)?

Explanation / Answer

The final reference point is 3/5 of the length below the launch point
so the PE change is mgh = (3/5)(4.6)(9.8)(2.03) = 54.90 joules

that is all converted to KE, so
KE = ½mV²
54.90 = ½*4.6*V²
V = 4.88 m/s

A simplified version, since m cancels, is
(3/5)mgh = ½mV²
V = [ (6/5)gh = 4.88m/s

2)

The net effect is that the mass has fallen through a height of (3/5) L
Hence v^2 = 2g h = 2g ((3/5) L
Now it tends to rotate above an arc of a circle of radius L/5
Hence mv^2/r = 2mg ((3/5) L / (L/5) = 6mg.
Tension = 6mg – mg = 5mg = 5*4.6*9.81 = 279.585 = 225.63 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.