Newton's Second Law in Angular Form — Question 5

Question 5

A radial constraint force cannot change axial angular momentum

A $0.50\,\mathrm{kg}$ bead moves on a fixed horizontal circle of radius $0.40\,\mathrm m$. A tangential force of constant magnitude $0.20\,\mathrm N$ acts counterclockwise. The bead starts at rest; all other horizontal forces are radial.

1. Find angular momentum and speed after $3\,\mathrm s$.

2. Find the radial constraint force magnitude at that time and explain its zero torque.

Question 5 - Solution

A radial constraint force cannot change axial angular momentum

(a) Derivation and calculation

The radial forces have zero moment about the circle center. The constant tangential torque is $$\tau_z=RF=.40(.20)=.0800\ \mathrm{N\,m}.$$ Starting from rest, $$\boxed{L_z(3)=\tau_z t=.0800(3)=.240\ \mathrm{kg\,m^2/s}},$$ $$\boxed{v=\frac{L_z}{mR}=\frac{.240}{.50(.40)}=1.20\ \mathrm{m/s}}.$$

(b) Result and interpretation

The radial constraint force supplies centripetal acceleration: $$\boxed{F_r=\frac{mv^2}{R}=\frac{.50(1.20)^2}{.40}=1.80\ \mathrm N\quad\text{inward}}.$$ Its torque is zero because $$\vec F_r\parallel-\vec r\Rightarrow\vec r\times\vec F_r=\vec0.$$ Force magnitude alone does not determine angular-momentum change; direction and lever arm matter.

Physical check

The radial force is nine times the tangential force but contributes none of the torque about the center.