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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,

$a_1=r{蝇}^2$

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

$r{蝇}^2=20\text{m}/{\text{s}}^2$

The centripetal acceleration of the second dust is,

$a_2=r_2{蝇}^2\left(\text{I}\right)$

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 $a_2$.

$a_2=\left(2r\right){蝇}^2=2\left(r{蝇}^2\right)$

Substitute $20\text{m}/{\text{s}}^2$for $r{蝇}^2$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,

$a_3=r{蝇}_3^2$

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

${蝇}_3=2蝇$Substitute $2{蝇}_1$for ${蝇}_3$in the equation (II) to find $a_3$.

$a_3=r{\left(2蝇\right)}^2=4\left(r{蝇}^2\right)$

Substitute$20\text{m}/{\text{s}}^2$for $r{蝇}^2$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$80\text{m}/{\text{s}}^2$.