physics! energy and work?

Question

In a circus act, a 60 kg trapeze artist starts from rest with the 5 m trapeze rope horizontal. What is the tension in the rope when it is vertical?

Please help! I have tried a bunch of things but I can't get the right answer!

Answer 1

Tension = weight + centripetal force
Weight = mass X Gravity acceleration
=60X9.8
=588N.

Centripeatl force = mV^2/r
G.P.E=K.E a bottom
mass X garvity acc. X hight = mass X velocity^2 X 0.5
60X9.8X5=1/2X60XV^2
V^2=98

Centripeatl force = mV^2/r
(60)(98)/5=1176 N.
Tension = 1176 + 588
=1764 N.

Tension = 1764 N.

Answer 2

This is a conservation of energy problem. So do an energy audit.

TE = PE + KE; total energy consists of potential and kinetic energy. TE will be constant no matter where the trapeze is in its swing.

Horizontal: TE = PE = mgh = potential energy; where m = 60 kg, g = 9.81 m/sec^2, and h = 5 meters = l the length of the rope. h = 0 when at the bottom on the swing, but l is still = 5 meters. KE = 0 because the trapeze is momentarily stationary.

Bottom: TE = KE = 1/2 mv^2 = kinetic energy; where v = velocity at the bottom of the swing and PE = 0 because h = 0 at that point.

From the conservation of energy: PE = mgh = 1/2 mv^2 = KE; then v^2 = 2gh. Now we can find the tension T.

T = W + C = mg + mv^2/l = m(g + v^2/l); where the tension on the rope equals the weight (W = mg) of the artist PLUS the cetripetal force (C = mv^2/l) due to the velocity v. As v^2 = 2gh, from above, we can write T = m(g + 2gh/l) = mg(1 + 2h/l) = 3mg; where m = 60 kg, g = 9.81 m/sec^2, h = l = 5 m so these cancel out. You can do the math.

The physics is this. TE, total energy, is constant no matter where in the swing the trapeze might be. So pick the two extremes (horizontal and vertical) and set the two TE's equal. Tension is the sum of the weight and the centripetal force because (W + C - T) = 0 = ma, which is clear since the artist (m) is not accelerating (a) along the length of the rope.

PS: Those answers who say the tension is three times the weight of the trapeze artist are correct. The others are not. This is a remarkable result. Whenever h = l, the tension will always be T = 3W = 3mg. For example, if the length of the trapeze rope were 100 meters and the trapezist started out from the horizontal with a taut rope, the tension would still be T = 3mg at the bottom of the swing. Who could've guessed?

Answer 3

okay, well firstly we know part of the tension is due to the artists weight so we calculate this
F(weight) = mg = 60 x 9.81
= 588.6 N

because the trapeze artist performs part of a rotation the rest of the tension is produced by the centripetal force acting towards where the rope is fixed,

Fc = mv^2/r but we need the trapeze artists tangential speed to solve for this, we could use conservation of energy to obtain v at the bottom of the rotation, when the rope is vertical. Do this, then use v, and centripetal force to find total tension.

total tension is
T = F(weight) + Fc

Answer 4

He is going to swing back and forth like a pendulum but will finally come to rest as long as he doesn't use his body to maintain the oscillations. Cetripedal acceleration plus gravitational force will be
the tension on the rope.
When the rope is vertical, the change in PE is = to change in KE:
mg*5=.5mv^2 5g=.5v^2
v^2=10g and v=sqrt[10g]=9.9m/s
The centripedal acceleration then is
mv^2/r=60v^2/5=12v^2=12*10g
=120g
The weight also adds to the tension so the
Total tension=[60+120]g=1764N
Hope this helps you

Answer 5

Conservation of energy:
m g L = (1/2) m V^2 ---> V^2 = 2 g L.
T - m g = m V^2/L ---> T = 3 m g = 1764 N

Answer 6

if you pull on a rope it only has the tension in it equal to your weight. but if you swing on a rope it has your weight plus centrepital force on it. get the formula from your book that tells the formula for centrepital force in terms of velocity v.

here is how to get velocity 0.5m(v squared)=mgh
solve for v. g is 9.8 m/s^2. h is 5 meters. just divide m out of both sides. m is 60.

now the tension in the rope is both weight plus centrepital added together. weight is 60x9.8
i think centripital is (60)x(v squared)/r or something.

Answer 7

dU+dKE=0
-mg*5-0=-1/mv^2
g*5=1/2v^2
v^2=100
now..
centripetal acc is v^2/r in upward direction ...
t-mg=mv^2/r
t=600+60*100/5
t=600+1200
t=1800N