How Do You Solve Complex Pendulum Problems Involving Tension and Angles?

[skivail]

Can anybody do these problems, and if you can, show and explain to me exactly how you did them?



The problems involve a pendulum bob of mass m that’s attached to a massless string of length l.



1. At what angle from the vertical should the bob be released so that the tension in the string at the bottom of the swing would be 1.8 times the pendulum’s weight?



2. Now, with the bob in #1 having swung through the vertical position, at what angle beyond the vertical does the string tension become exactly equal to the bob’s weight?

 

[Leong]

At the bottom,
T-mg=\frac{mv^2}{l}
T=1.8 mg; then
v^2=0.8gl
v is the speed of the pendulum at the bottom.
Law of conservation of energy :
mgh=\frac{1}{2}v^2
h is the height of the pendulum relative to the bottom before it is released.
h=0.4l
Trig formula :
cos\theta=\frac{l-0.4l}{l}
\theta=53.1^0
For question #2, use
1. Newton's 2nd law and the centripedal force formula.
2. Use the law of conservation of energy.
Similar to question #1.

 

[skivail]

Thanks.

Is there anyway that you could work out part 2 so i could check my work?

 

[Leong]

Centripedal force & Newton's 2nd Law & Law of conservation of energy

At that position :
T-mgcos\alpha=\frac{mv^2}{l}
mg(1-cos\alpha)=\frac{mv^2}{l}...(1)
v is the speed of the pendulum at that position.
Law of conservation of energy :
Pendulum's initial energy = Pendulum's energy at that position
mgh=\frac{1}{2}mv^2+mgH
H is the height of the pendulum relative to the bottom at that position.
h is the initial height of the pendulum calculated in question #1.
H=l-lcos\alpha
=l(1-cos\alpha)
Then
0.4gl=\frac{1}{2}v^2+gl(1-cos\alpha)
v^2=0.8gl-2gl(1-cos\alpha)...(2)
(2) into (1) :
g(1-cos\alpha)=\frac{0.8gl-2gl(1-cos\alpha)}{l}
1-cos\alpha=0.8-2+2cos\alpha
\alpha=42.8^0