Uniform Circular Motion — Question 4

Question 4

A conical pendulum couples geometry to force balance

Take $g=10$ m/s$^2$. A 2 kg bob on a light 2 m cord executes a steady horizontal circle. The cord makes $60^\circ$ with the vertical. The support is fixed and air resistance is negligible.

1. Find the circular radius and cord tension.

2. Find the speed and period.

3. Derive the period in terms of cord length L and angle, and explain why using L as the radius is incorrect.

Question 4 - Solution

A conical pendulum couples geometry to force balance

(a) Use geometry and vertical balance.

The circle radius is the horizontal projection of the cord: $$\boxed{r=L\sin60^\circ=\sqrt3\ \mathrm m}.$$ Vertical force balance gives $$\boxed{T=\frac{mg}{\cos60^\circ}=\frac{2(10)}{.5}=40\ \mathrm N}.$$

(b) Use the horizontal force component.

Use the inward horizontal component of tension: $$T\sin\theta=\frac{mv^2}{r}\Longrightarrow v^2=rg\tan\theta=30\ \mathrm{m^2/s^2}.$$ $$\boxed{v=\sqrt{30}=5.477\ \mathrm{m/s}},\qquad \boxed{\mathcal T=\frac{2\pi\sqrt3}{\sqrt{30}}=\frac{2\pi}{\sqrt{10}}=1.987\ \mathrm s}.$$

(c) Eliminate the speed carefully.

Eliminate speed and then use the geometry: $$\mathcal T=2\pi\sqrt{\frac{r}{g\tan\theta}} =2\pi\sqrt{\frac{L\sin\theta}{g\tan\theta}}.$$ $$\boxed{\mathcal T=2\pi\sqrt{\frac{L\cos\theta}{g}}}.$$ Using $L$ as the circle radius ignores its vertical projection. This expression describes steady conical motion with a nonzero angle.