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In a Children'S Park a Heavy Rod is Pivoted at the Centre and is Made to Rotate About the Pivot So that the Rod Always Remains Horizontal. Two Kids Hold the Rod Near the Ends - Physics

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प्रश्न

In a children's park a heavy rod is pivoted at the centre and is made to rotate about the pivot so that the rod always remains horizontal. Two kids hold the rod near the ends and thus rotate with the rod (In the following figure). Let the mass of each kid be 15 kg, the distance between the points of the rod where the two kids hold it be 3.0 m and suppose that the rod rotates at the rate of 20 revolutions per minute. Find the force of friction exerted by the rod on one of the kids. 

बेरीज

उत्तर

Given :

\[\text {Frequency of rod }= n = 20 \text{ rev} \text{ per } \min\]

\[ \Rightarrow \text{ n} = \frac{20}{60} = \frac{1}{3}\text{rev/s}\]

Therefore, we have :
angular velocity of rod ,

\[\text { Angular velocity of rod }= \omega = 2\pi n = \frac{2\pi}{3}\text{rad/s}\]

Mass of each kid = \[\text{m = 15 kg}\]

Radius = \[r = \frac{3}{2} = 1 . 5 \text{m}\]

\[\therefore \text { Frictional force }= F = \text{mr }\omega^2 \]

\[ \Rightarrow F = 15 \times (1 . 5) \times \frac{(2\pi )^2}{9}\]

\[ = 5 \times (0 . 5) \times 4 \pi^2 = 10 \pi^2 N\]

Thus, the force of frictional on one of the kids is 10 \[\pi\]

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पाठ 7: Circular Motion - Exercise [पृष्ठ ११५]

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एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
पाठ 7 Circular Motion
Exercise | Q 23 | पृष्ठ ११५

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