Question
A mass hangs on the end of a massless rope. The pendulum is held horizontal and released from rest. When the mass reaches the bottom of its path it is moving at a speed v = 2.3 m/s and the tension in the rope is T = 22.4 N.
1. What is the mass?
2. If the maximum mass that can be used before the rope breaks is mmax = 1.72 kg, what is the maximum tension the rope can withstand? (Assuming that the mass is still released from the horizontal.)
1. What is the mass?
2. If the maximum mass that can be used before the rope breaks is mmax = 1.72 kg, what is the maximum tension the rope can withstand? (Assuming that the mass is still released from the horizontal.)
Answers
1.What is the mass
find the length of the rope
v^2/(2*g)=L
find the mass
(L*T)/(v^2+L*g)=m
find the length of the rope
v^2/(2*g)=L
find the mass
(L*T)/(v^2+L*g)=m
Second equation is correct,
First equation is (1/2mv^2)/mg => (1/2v^2)/g
First equation is (1/2mv^2)/mg => (1/2v^2)/g
(1/2v^2)/g is the same as v^2/(2g) so both work!