Question

A uniform wheel of moment of inertia 'I' is pivoted on a horizontal axis through its centre so that its plane is vertical as shown in the figure. A small mass 'm' is stuck on the rim of the wheel as shown. The angular acceleration of the wheel when mass is at point A is ${\alpha }_A$ and when mass is at point B is ${\alpha }_B$. Then,

• A. $\frac{{\alpha }_B}{{\alpha }_A}=\sin \theta$
• B. $\frac{{\alpha }_B}{{\alpha }_A}={\sin }^2\theta$
• C. $\frac{{\alpha }_B}{{\alpha }_A}=\cos \theta$
• D. $\frac{{\alpha }_B}{{\alpha }_A}={\cos }^2\theta$

Answer 1

The correct option is C $\frac{{\alpha }_B}{{\alpha }_A}=\cos \theta$

The torque due to the weight of m at A is ${\tau }_A$ and that when it is at B is ${\tau }_B$
$\Rightarrow \frac{{\tau }_B}{{\tau }_A}=\frac{I{\alpha }_B}{I{\alpha }_A}=\frac{b\cos \theta }{b}=\cos \theta .$