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Assoli18 [71]
2 years ago
7

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?
Physics
1 answer:
Bond [772]2 years ago
6 0

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}

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Answer:

U(0.0m) = 0J

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The gravitational potential energy is given by the following formula:

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