91Ó°ÊÓ

The centripetal acceleration of a speck of dust on a spinning DVD is $20\text{m}/{\text{s}}^2$, the radius of rotation of the different dust is twice as far from the centre of the disk.

The centripetal acceleration of the first dust is,

$a_1=r{\mathrm{Ó\neg }}^2$

Substitute $20\text{m}/{\text{s}}^2$for $a_1$in the above equation to find $r{\mathrm{Ó\neg }}^2$.

$r{\mathrm{Ó\neg }}^2=20\text{m}/{\text{s}}^2$

The centripetal acceleration of the second dust is,

$a_2=r_2{\mathrm{Ó\neg }}^2\left(\text{I}\right)$

It is given that the radius of rotation of the different dust is twice as far from the centre of the disk. Thus $r=2r$

Substitute 2r for$r_2$in the equation (I) to find $a_2$.

$a_2=\left(2r\right){\mathrm{Ó\neg }}^2=2\left(r{\mathrm{Ó\neg }}^2\right)$

Substitute $20\text{m}/{\text{s}}^2$for $r{\mathrm{Ó\neg }}^2$in the above equation to find $a_2$.

$a_2=2\left(20\text{m}/{\text{s}}^2\right)=40\text{m}/{\text{s}}^2$

Thus, the acceleration of the second speck of dust is $40\text{m}/{\text{s}}^2$.

The centripetal acceleration of a speck of dust on a spinning DVD is $20\text{m}/{\text{s}}^2$.

The centripetal acceleration of the first dust is,

$a_3=r{\mathrm{Ó\neg }}_3^2$

It is given that the angular acceleration of the dust is doubled from the initial. Thus,

${\mathrm{Ó\neg }}_3=2\mathrm{Ó\neg }$Substitute $2{\mathrm{Ó\neg }}_1$for ${\mathrm{Ó\neg }}_3$in the equation (II) to find $a_3$.

$a_3=r{\left(2\mathrm{Ó\neg }\right)}^2=4\left(r{\mathrm{Ó\neg }}^2\right)$

Substitute$20\text{m}/{\text{s}}^2$for $r{\mathrm{Ó\neg }}^2$in the above equation to find $a_3$.

$a_3=4\left(20\text{m}/{\text{s}}^2\right)=80\text{m}/{\text{s}}^2$

Thus, the acceleration of the first speck of dust when the disk's angular velocity was doubled is$80\text{m}/{\text{s}}^2$.