A mass m = 6.5 kg hangs on the end of a massless rope L = 2.01 m long. The pendu

Question

A mass m = 6.5 kg hangs on the end of a massless rope L = 2.01 m long. The pendulum is held horizontal and released from rest.

1)

How fast is the mass moving at the bottom of its path?

m/s

2)

What is the magnitude of the tension in the string at the bottom of the path?

N

3)

If the maximum tension the string can take without breaking is Tmax = 541 N, what is the maximum mass that can be used? (Assuming that the mass is still released from the horizontal and swings down to its lowest point.)

kg

4)

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)?

m/s

5)

Using the original mass of m = 6.5 kg, what is the magnitude of the tension in the string at the top of the new path (directly above the peg)

please answer #5

Explanation / Answer

1) At the start of the motion, the total energy is potential energy.

Initial Potential energy, Ui = mgh

At the bottom of the path, the total energy is kinetic energy. Let v be the velocity at the bottom of the path.

Kinetic energy, K.E = (1/2)mv^2

Then by conservation of energy,

mgh = (1/2)mv^2

=> v = sqrt(2gh) = sqrt(2*9.81*2.01) = 6.28m/s

2) The net force acting on the pendulum at the bottom is is

centripetal force = weight - tension in the string

i.e mv2/L = T-mg

=> T = m(g+v2/L) = 6.5(9.81+(6.28)2/2.01) = 191.29N

3) If T = 541N, then, m = 541/(g+v2/L) = 541/29.43 = 18.38Kg

This is the max mass that can be supported.

4) The bob is now at a point (3/5*2.17) = 1.206 m
From rest it has gained a speed v.
v^2 = 2gh = 2*9.81*1.206
v = 4.86m/s

5) Now the pendulum tends to rotate above an arc of a circle of radius L/5
Hence mv^2/r = m[2g(3/5)L / (L/5) = 6mg
Hence, Tension = 6mg – mg = 5mg = 5*6.5*9.81 = 318.825N

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