1) A mass m = 4.5 kg hangs on the end of a massless rope L = 2.01 m long. The pe

Question

1)

A mass m = 4.5 kg hangs on the end of a massless rope L = 2.01 m long. The pendulum is held horizontal and released from rest. How fast is the mass moving at the bottom of its path? What is the magnitude of the tension in the string at the bottom of the path? If the maximum tension the string can take without breaking is Tmax = 388 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.) 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)? Using the original mass of m = 4.5 kg, what is the magnitude of the tension in the string at the top of the new path (directly above the peg)?

Explanation / Answer

Potential energy:
Ep=mgh with h=L

For any point of the pendulum path:
Ep= mgL(1-cos(q)) => With q= angle from the resting possition(horizontal).

Kinetic energy (rotation):
Ek= 1/2 I W^2 => With I=Inertia momentum and W= angular speed

If the size relation between the mass and the rope can be neglected. The mass volume and shape do not affect the motion.

I = m r^2 => With r = L

Then Ek= 1/2 m L^2 W^2

At any point of the pendulum path, energies must be balanced:

Ep= Ek

mgL(1-cos(q)) = 1/2 m L^2 W^2

W^2= (2g(1-cos(q))/L

W= SQRT{(2g(1-cos(q))/L]}

For q=0: The pendulum is held up, in the horizontal possition: cos(0)=1; W=0
For q=PI/2; The pendulum is at its vertical possition: cos(PI/2)=0; W=max.

W= SQRT{(2g/L]} => W= SQRT(2 x 9.81/20.1) =0.976 rads/sec

Linear speed V: V= W x L = 19.617 m/s

Centrifugal force:

F= mv^2/L; F= 4.5 x 19.617^2/20.1 = 86,155 Kg m^2/s^2/m = Kg m/s^2 = Nw
Max Mass:

m= LxF/V^2: F= 388 Nw
m= 20.1 x 388/18.155^2
m= 23.66 Kg

initial PE = 4.5 * 9.81 * 2.01
final KE = 0.5 * 4.5 * u^2
4.5 * 9.81 * 2.01 = 0.5 * 4.5 * u^2
u = 6.28m/s
tension in the string at the bottom of the path
= mg + mu^2/r
= 4.5 * 9.81 + (4.5 * 6.28^2 / 2.01)

=132.44

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