A mass m = 4.7 kg hangs on the end of a massless rope L = 2.05 m long. The
pendulum is held horizontal and released from rest.
1) How fast is the mass moving at the bottom of its path?
2) What is the magnitude of the tension in the string at the bottom of the
path?
3) If the maximum tension the string can take without breaking is Tmax =
382 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.)
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)?

Respuesta :

Answer:

1) v=sqrt (2gh)

v = sqrt ( 2 x 9.8 x 2.05) = 6.3 m/s

2) T=mg+mv^2/r

T = 4.7 x 9.8 + 4.7 x 6.3^2 / 2.05

T = 36.06 + 90.99 = 127 N

3. m= T/(g+v^2/r)

m = 382 / ( 9.8 + 6.3^2 / 2.05)

m = 382 / 29.16 = 13.1

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