Conservation of Mechanical Energy — Question 3

Question 3

Pendulum amplitude inferred from a bottom speed

Neglect friction and air resistance, take $g=10$ m/s$^2$, and treat all stated guides as fixed. A 0.40 kg bob on a massless 1.5 m cord passes the bottom with speed $\sqrt{15}$ m/s. The pivot is fixed.

1. Find the maximum angle from downward vertical.

2. Find speed at an angle of $30^\circ$.

3. Check that cord tension is positive at the turning point.

Question 3 - Solution

Pendulum amplitude inferred from a bottom speed

(a) Compare bottom kinetic energy with the height gain.

The turning-point speed vanishes: $$\tfrac12mv_0^2=mgL(1-\cos\theta_{\max}),\quad1-\cos\theta_{\max}=\frac{15}{2(10)(1.5)}=\frac12.$$ $$\boxed{\theta_{\max}=60^\circ}.$$

(b) Subtract the partial gravitational rise.

At $30^\circ$, $$v^2=15-30(1-\cos30^\circ)=15(\sqrt3-1),$$ $$\boxed{v=\sqrt{15(\sqrt3-1)}=3.314\ \mathrm{m/s}}.$$

(c) Use the radial equation at zero speed.

The radial equation toward the pivot is $$T-mg\cos\theta=\frac{mv^2}{L}.$$ At the turn, $\boxed{T=.40(10)\cos60^\circ=2\ \mathrm N>0}$. Throughout $|\theta|\leq60^\circ$, both terms in $T=mv^2/L+mg\cos\theta$ are nonnegative and the second is positive.