Question

A 16-cm-long straight line connects the center of a turntable to its edge. The turntable rotates counter-clockwise at 45 rpm. A bug crawls along the line from the turntable’s center at 3.5 cm s relative to the turntable. Assume it is initially moving in the positive x direction. At the moment the bug gets to the edge, what are the x and y components of the velocity of the bug?

Answer 1

Answer:

$\mathbf{V_x = 3.25 \ cm/s}$

$\mathbf{V_y = 1.29\ cm/s}$

Explanation:

Given that:

The radius of the table r = 16 cm  = 0.16 m

The angular velocity = 45 rpm

= $45 \times \dfrac{1}{60}(2 \pi)$

= 4.71 rad/s

However, the relative velocity of the bug with turntable is:

v = 3.5 cm/s = 0.035 m/s

Thus, the time taken to reach the bug to the end is:

$t = \dfrac{r}{v}$

$t = \dfrac{0.16}{0.035}$

t = 4.571s

So the angle made by the radius r  with the horizontal during the time the bug gets to the end is:

$\theta = \omega t$

$\theta = 4.712 \times 4.571$

$\theta = 21.54^0$

Now, the velocity components of the bug with respect to the table is:

$V_x = Vcos \theta$

$V_x = 0.035 \times cos (21.54^0)$

$V_x = 0.0325 \ m/s$

$\text {V_x = 3.25 \ cm/s}$$\mathbf{V_x = 3.25 \ cm/s}$

Also, for the vertical component of the velocity $V_y$

$V_y = V sin \theta$

$V_y = 0.035 \times sin (21.54^0)$

$V_y = 0.0129\ m/s$

$\mathbf{V_y = 1.29\ cm/s}$