The Yo-Yo — Question 6

Question 6

An ascending yo-yo can accelerate downward before reversing

A yo-yo with $I=3Mr^2$ moves upward at $1.0\,\mathrm{m/s}$ while winding onto a fixed cord. Ignore losses and use $g=10$. The no-slip constraint remains valid through the turning point.

1. Find its acceleration, time to the highest point and rise.

2. Find tension at the highest point; does zero speed imply zero torque?

Question 6 - Solution

An ascending yo-yo can accelerate downward before reversing

(a) Derivation and calculation

The downward acceleration does not depend on the instantaneous direction of motion: $$\boxed{a=\frac{g}{1+I/(Mr^2)}=10/4=2.50\ \mathrm{m/s^2}}.$$ With initial downward velocity $v_0=-1.0$ m/s, $$0=v_0+at\Rightarrow\boxed{t_{\mathrm{top}}=.400\ \mathrm s},$$ $$0=v_0^2+2a\Delta y\Rightarrow\boxed{\text{rise}=-\Delta y=.200\ \mathrm m}.$$

(b) Result and interpretation

At the highest point, $$\boxed{T=M(g-a)=M(7.50\ \mathrm{m/s^2})},$$ $$\tau=Tr\ne0,\qquad\alpha=a/r=2.50/r\ne0.$$ Although $v=\omega=0$ at that instant, the tension and accelerations remain finite and reverse the winding motion into unwinding descent.

Physical check

The turning point changes the velocity signs, not the constant downward acceleration.