Problem

A point on Blu-ray disc is a distance R/4  from the axis of rotation. How far from the axi...

A point on Blu-ray disc is a distance R/4  from the axis of rotation. How far from the axis of rotation is a second point that has, at any instant, a linear velocity twice that of the first point?

a) R/16

b) R/8

c) R/2

d) R

Step-by-Step Solution

Solution 1

The distance from the rotational axis depends on the angular speed and the linear speed. Since, the linear velocity can be obtained by taking the product of the angular velocity and distance from the rotational axis to the point.

Let the initial linear velocity of Blu-ray disc is v₁ The final linear velocity of Blu-ray disc, v₂=2v₁

The linear velocity of the Blu-ray disc at \(\frac{R}{4}\) distance is given by the equation \(v_{1}=\left(\frac{R}{4}\right) \omega \ldots \ldots\) (1)

The linear velocity of the Blu-ray disc at the second point is given by the equation \(v_{2}=x \omega \ldots \ldots\) (2)

Since, the angular velocity of the disc is constant.

Using equations (1) and (2), we get the unknown distance ' \(x\) '

$$ \begin{aligned} \frac{v_{1}}{v_{2}} &=\frac{\left(\frac{R}{4}\right) \omega}{x \omega} \ldots \ldots(3) \\ x &=\frac{R v_{2}}{4 v_{1}} \end{aligned} $$

Now, substituting the numerical values in the equation (3), we get

$$ \begin{aligned} x &=\frac{R v_{2}}{4 v_{1}} \\ &=\frac{R\left(2 v_{1}\right)}{4 v_{1}} \\ &=\frac{R}{2} \end{aligned} $$

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