Relating the Linear and Angular Variables — Question 3

Question 3

An open belt shares rim speed rather than angular speed

Two fixed pulleys of radii $0.10$ and $0.25\,\mathrm m$ are connected by a taut uncrossed belt with no slip. The smaller rotates counterclockwise at $8\,\mathrm{rad/s}$ and accelerates at $2\,\mathrm{rad/s^2}$.

1. Find the larger pulley's angular velocity and acceleration.

2. Compare inward accelerations of belt-contact points on the two rims.

Question 3 - Solution

An open belt shares rim speed rather than angular speed

(a) Derivation and calculation

No slip imposes equal belt and rim speeds; an uncrossed belt preserves the rotation sense: $$R_1\omega_1=R_2\omega_2\Rightarrow \boxed{\omega_2=\frac{0.10}{0.25}(8)=3.20\ \mathrm{rad/s}}.$$ $$R_1\alpha_1=R_2\alpha_2\Rightarrow \boxed{\alpha_2=\frac{0.10}{0.25}(2)=0.80\ \mathrm{rad/s^2}}.$$

(b) Result and interpretation

The shared instantaneous rim speed is $v=0.10(8)=0.80$ m/s. Thus $$\boxed{a_{{\mathrm{in}},1}=\frac{v^2}{R_1}=\frac{0.64}{0.10}=6.40\ \mathrm{m/s^2}},\qquad \boxed{a_{{\mathrm{in}},2}=\frac{0.64}{0.25}=2.56\ \mathrm{m/s^2}}.$$ Equal tangential speeds and tangential accelerations do not require equal curvature or inward acceleration.

Physical check

The larger pulley has the smaller angular speed and the smaller inward acceleration.