2013: Problems 19-21

Topic: Dynamics
Concepts: Circular motion, Newton’s laws


Solution:

Since the pendulum is undergoing circular motion, we have

Tmgcosθ=mv2LT-mg\cos\theta=\frac{mv^2}{L}

so the tension is

T=mgcosθ+mv2LT=mg\cos\theta+\frac{mv^2}{L}

At the bottom θ=0\theta=0, cosθ\cos\theta is maximized since cos0=1\cos\,0=1 while vv is also maximized by conservation of energy (pendulum is at lowest point of swing). Thus, both terms are maximized so the tension is largest at θ=0\theta=0. Hence, the answer is B.

Topic: Energy
Concepts: Circular motion, Conservation of energy


Solution:

We found in the previous problem that the tension is maximized when the pendulum is at the bottom. We have

Tmg=mv2LT-mg=\frac{mv^2}{L}

Since the pendulum amplitude is θ0\theta_0, the pendulum has fallen a height

h=L(1cosθ0)h=L(1-\cos\theta_0)

By conservation of energy,

mgh=12mv2mgh=\frac{1}{2}mv^2
v2=2gh=2gL(1cosθ0)v^2=2gh=2gL(1-\cos\theta_0)

Substituting this into the force equation,

Tmg=2mg(1cosθ0)T-mg=2mg(1-\cos\theta_0)
T=mg(32cosθ0)T=mg(3-2\cos\theta_0)

so the answer is E.

Topic: Oscillatory Motion
Concepts: Dimensional analysis, Simple pendulum


Solution:

By dimensional analysis, the period of a simple pendulum is given by

T=f(θ)LgT=f(\theta)\sqrt{\frac{L}{g}}

We are given

T0=f(θ0)LgT_0=f(\theta_0)\sqrt{\frac{L}{g}}

For the new pendulum of length 4L4L with the same amplitude θ0\theta_0,

T=f(θ0)4Lg=2T0T=f(\theta_0)\sqrt{\frac{4L}{g}}=2T_0

so the answer is A.

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