Answer
(a) $0.82m/s$ (b) $9.9m/s^2$ (c) The velocity would decrease (by half) and the acceleration would decrease (by a fourth).
Work Step by Step
(a) The period of rotating is equal to $T=\frac{2\pi}{\omega}$, so $$\omega=\frac{2\pi}{T}$$ Since $v=r\omega$ and $\omega=\frac{v}{r}$, the expression becomes $$\frac{v}{r}=\frac{2\pi}{T}$$ $$v=\frac{2\pi r}{T}$$ Substituting known values of $T=0.52s$ and $r=6.8cm=0.068m$ yields a velocity of $$v=\frac{2\pi(0.068m)}{0.52s}=0.82m/s$$ (b) The centripetal acceleration of a point on a rotating body is equal to $$a=\frac{v^2}{r}$$ Substituting known values of $v=0.82m/s$ and $r=0.068m$ yields an acceleration of $$a=\frac{(0.82m/s)^2}{0.068m}=9.9m/s^2$$ (c) If the period is doubled, the velocity of the clay must decrease (by half). If velocity decreases and the radius remains the same, the centripetal acceleration will also decrease (by a fourth).
