iPad * 48%. 12:42 AM flipitphysics.com Pendulum 2 Loop the Loop IE Block Spring
ID: 3281302 • Letter: I
Question
iPad * 48%. 12:42 AM flipitphysics.com Pendulum 2 Loop the Loop IE Block Spring Incline A mass hangs on the end of a massless rope. The pendulum is held horipontal and released from rest. When the mass reaches the battom of ies path i h moving at a speed v-2.1m/sand the tension in the rope is T-188 1) How long s the rope 2) What is the mass? 3) If the mamum mass that can be used before the rope breaks is a-1.34 kg. what is the maximum tension the rope can wihstand (Assuming that the mass is still released from the horizontal. ) 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 wrapo around the pe How fast is the mass moving when t is at the same vertical height as the peg Idirectly to the right of the pegi 5) Return to the original mass. What is the tension in the sering at the same vertical heigh as the peg irectly to the right of the pegi? l Below is some space to write notes on this problemExplanation / Answer
1) Let L be the length of the rope
when mass reachs the bottom most position it has lost a height of l and change in Pot. energy = mgl
conserving the total mechanical energy
KE of the mass at the bottom =mv2/2 = mgl
speed of the mass v = sqrt(2gL) = 2.1 m/s
L = 0.225 m
2. Tension in the rope = 18.8 N
when the mass is at its bottom, the only forces acting on it are
1. its weight mg
2. tension in the rope T
3. centrifugal force mv2/r
in vertical direction the mass is equilibrium and the forces equal
T = mg + mv2/R= 18.8 N
mass m (9.8 + 2.12/0.225)= 18.8
m = 0.64 kg
3. max mass M = 1.34 kg
max tension Tmax = 1.34 *9.8 + 1.34*2.12/0.225 = 39.40 N
4. the peg is 4/5L down = 0.225*4/5 = 0.18 m
when the mass raises to the level of the peg increase in height h = 0.225 -0.18 = 0.045 m
it will loose KE , equal to the pot.energy = 0.64*9.8*0.045 = 0.282 J
initial KE = 0.5*0.64*2.1 = 1.41 J
final KE = 1.41- 0.28 = 1.13 J
speed = sqrt (2*1.13/0.64) = 1.88 m/s
5. when the mass is to the right of the peg speed v = 1.88 m/s
in this position
The forces acting in the horizontal direction are
1. tension in the rope
2. centrifugal force mv2/r
the mass is waraping arroung the ped hence free length of the string r = 0.045 m
tension T = 0.64 *1.882/0.045 = 50.27 N
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