Rotation Ques 102
- A uniform circular disc of mass $1.5 kg$ 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 $\operatorname{rad~s}^{-1}$ is (2012)
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Answer:
Correct Answer: 102.(2)
Solution:
- Angular impulse $=$ change in angular momentum
$$ \begin{aligned} & \therefore \int \tau d t=I \omega \\ & \Rightarrow \quad \omega=\frac{\int \tau d t}{I}=\frac{\int _0^{t} 3 F \sin 30^{\circ} R d t}{I} \end{aligned} $$
Substituting the values, we have
$$ \omega=\frac{3(0.5)(0.5)(0.5)(1)}{\frac{1.5(0.5)^{2}}{2}}=2 rad / s $$
BETA
