Answer
(a) $v=0.30m/s$ (b) $1.5m/s^2$ (c) $0.15m/s$ (d) $0.75m/s^2$
Work Step by Step
(a) We know that
$v=r\omega$
$v=(0.06m)(5.05rad/s)$
$v=0.30m/s$
(b) The centripetal acceleration is given as:
$a_{cp}=r\omega^2$
We plug in the known values to obtain:
$a_{cp}=(0.06m)(5.05rad/s)^2=1.5m/s^2$
(c) When the radius is reduced to half, the linear speed is also reduced to half
$v=\frac{0.303m/s}{2}=0.15m/s$
As the centripetal acceleration is directly proportional to the radius of the CD, therefore, when the radius is reduced to half then the centripetal acceleration is also reduced to half.
$\implies a_{cp}=\frac{1.53m/s^2}{2}$
$\implies a_{cp}=0.75m/s^2$
