A uniform rod of length 'l' is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed ω  the rod makes an angle   θ with it (see figure). To find θ equate the rate of change of angular momentum (direction going into the paper)   m l 2 12 ω 2 s i n ? θ c o s ? θ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces   F H and   F V about the C M   . The value of θ is then such that:

 

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mi mathvariant="normal">c</mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">s</mi> <mo>⁡</mo> <mi>θ</mi> <mo>=</mo> <mfrac> <mrow> <mrow> <mi>g</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> <msup> <mrow> <mrow> <mi>ω</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> </span></p>
Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mi mathvariant="normal">c</mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">s</mi> <mo>⁡</mo> <mi>θ</mi> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>2</mn> <mi>g</mi> </mrow> </mrow> <mrow> <mrow> <mn>31</mn> <msup> <mrow> <mrow> <mi>ω</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> </span></p>
Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mi mathvariant="normal">c</mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">s</mi> <mo>⁡</mo> <mi>θ</mi> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>3</mn> <mi>g</mi> </mrow> </mrow> <mrow> <mrow> <mn>21</mn> <msup> <mrow> <mrow> <mi>ω</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> </span></p>
Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mi mathvariant="normal">c</mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">s</mi> <mo>⁡</mo> <mi>θ</mi> <mo>=</mo> <mfrac> <mrow> <mrow> <mi mathvariant="normal">g</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mi mathvariant="normal">l</mi> <msup> <mrow> <mrow> <mi>ω</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> </span></p>
18 Views|Posted 5 months ago
Asked by Shiksha User
1 Answer
A
5 months ago
Correct Option - 3
Detailed Solution:

Torque of centrifugal force about A

= d = d m ω 2 ( x s i n ? θ ) ( x c o s ? θ )

= 0 l ? m l d x ω 2 x 2 s i n ? θ c o s ? θ

= m l ω 2 s i n ? θ c o s ? θ x 3 3 0 l = m ω 2 l 2 s i n ? θ c o s ? θ 3

  τ m g = τ centrifugal   (about A)

m g l 2 s i n ? θ = m ω 2 l 2 s i n ? θ c o s ? θ 3

c o s ? θ = 3 g 2 l ω 2

 

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Physics System of Particles and Rotational Motion 2025

Physics System of Particles and Rotational Motion 2025

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