A mass m = 4.8 kg hangs on the end of a massless rope L = 2.07 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 = 393 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)?
5. Using the original mass of m = 4.8 kg, what is the magnitude of the tension in the string at the top of the new path (directly above the peg)?
3 Answers

1) PE = KE
mgh = ½mv²
v = √(2gh)
v = √(2(9.81)2.07)
v = 6.37 m/s
2) The string must support the weight of the mass plus create the centripetal acceleration.
T = m(v²/L + g)
T = 4.8(6.37²/2.07 + 9.81)
T = 141 N
3) m = Tmax/(v²/L + g)
m = 393/(6.37²/2.07 + 9.81)
m = 13.4 kg
4) kinetic energy will be equal the the change in potential energy from the start point
v = √(2g(3L/5)
v = √(2(9.81)(3(2.07)/5)
v = 4.94 m/s
5) Tension will be due to centripetal acceleration, but gravity supplies part of the acceleration
T = m(v²/(L/5)  g)
T = 4.8(4.94²/(2.07/5)  9.81)
T = 235 N
As a form of exchange, please remember to vote a "Best Answer" from among your results. This is not the same as "like" or "thumbs up", which are also nice and do boost points, but only after initial points are awarded for asker's "Best Answer"

This Site Might Help You.
RE:
Physics Pendulum problem!!?
A mass m = 4.8 kg hangs on the end of a massless rope L = 2.07 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...
Source(s): physics pendulum problem: https://shortly.im/G9MVe 
PE = (mgh) = (4.8 x 9.8 x 2.07) = 97.3728 Joules. This converts to KE at the bottom of the arc.
1) V at bottom = sqrt.(2KE/m) = sqrt.(194.7456/4.8) = 6.37m/sec.
2) Acceleration = (v^2/r) = (6.37^2/2.07) = 19.6m/sec^2.
Tension = (ma) = 4.8 x 19.6 = 94.08N.
3) The acceleration will be the same, 19.6m/sec^2.
Mass = (f/a) = 393/19.6 = 20kg. max.
4) (4/5 of 2.07m) = 1.242 metres. The new pendulum length is 0.828 metres.
KE = (mgh) = 4.8 x 9.8 x 1.242 = 58.42368 Joules (converted from GPE).
V at peg = sqrt.(2KE/m) = sqrt.(116.84736/4.8) = 4.934m/sec.
The short pendulum now gains PE of (mgh) = (4.8 x 9.8 x (0.828 x 2) = 27.898 Joules.
Its KE at the bottom was 58.42368 Joules, subtract 27.898 = KE of 30.5257 Joules at top of arc.
V = sqrt.(2KE/m) = sqrt.(61.05136/4.8) = 3.57m/sec.
5) Tension = (mv^2/r)  (mg) = ((4.8 x 3.57^2)/0.828)  (4.8 x 9.8) = 14.135N.