SATHEE: Rotation 5 Question 19

Rotation 5 Question 19

31. A uniform circular disc of mass $1.5 k g$ and radius $0.5 m$ is initially at rest on a horizontal frictionless surface. Three forces of equal magnitude $F = 0.5 N$ are applied simultaneously along the three sides of an equilateral triangle

$X Y Z$ with its vertices on the perimeter of the disc (see figure). One second after applying the forces, the angular speed of the disc in ${rad s}^{- 1}$ is (2012)

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Answer:

Correct Answer: 31. (2)

Solution:

Angular impulse $=$ change in angular momentum

$\begin{aligned} ∴ \int τ d t = I ω \\ \Rightarrow ω = \frac{\int τ d t}{I} = \frac{\int_{0}^{t} 3 F \sin 30^{\circ} R d t}{I} \end{aligned}$

Substituting the values, we have

$ω = \frac{3 (0.5) (0.5) (0.5) (1)}{\frac{1.5 (0.5)^{2}}{2}} = 2 r a d / s$

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Rotation 5 Question 19

31. A uniform circular disc of mass 1.5kg and radius 0.5m is initially at rest on a horizontal frictionless surface. Three forces of equal magnitude F=0.5N are applied simultaneously along the three sides of an equilateral triangle

Fill in the Blanks

Answer:

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31. A uniform circular disc of mass $1.5 k g$ and radius $0.5 m$ is initially at rest on a horizontal frictionless surface. Three forces of equal magnitude $F = 0.5 N$ are applied simultaneously along the three sides of an equilateral triangle