Linear Momentum — Question 3

Question 3

Constant momentum magnitude still permits force

A $0.20\,\mathrm{kg}$ bead moves at constant speed $5.0\,\mathrm{m/s}$ around a circle of radius $2.0\,\mathrm m$. At the rightmost point it moves upward.

1. Find its momentum and net force vectors at that point using rightward $+x$ and upward $+y$.

2. Find its momentum change over the next half revolution and explain why $d|\vec p|/dt=0$ does not imply zero force.

Question 3 - Solution

Constant momentum magnitude still permits force

(a) Derivation and calculation

At the rightmost point, velocity is upward and acceleration is inward: $$\boxed{\vec p=mv\hat y=(0,1.0)\ \mathrm{kg\,m/s}}.$$ $$\boxed{\vec F=-\frac{mv^2}{R}\hat x=-\frac{0.20(25)}{2}\hat x=(-2.5,0)\ \mathrm N}.$$

(b) Result and interpretation

After half a revolution the tangent reverses: $$\vec p_f=(0,-1),\qquad\boxed{\Delta\vec p=\vec p_f-\vec p_i=(0,-2)\ \mathrm{kg\,m/s}}.$$ Although $d|\vec p|/dt=0$, the vector derivative $d\vec p/dt$ is nonzero. Force changes the direction of momentum.

Physical check

The inward force is perpendicular to velocity and therefore does no work.