Question

A block of mass $m$ is placed at the top of a smooth wedge ABC. The wedge is rotated about an axis passing through C as shown in the figure. The value of angular speed $\omega$ such that the block at $A$ does not slip on the wedge is

• A. $\left(\sqrt{\frac{g\sin \theta }{l}}\right)\sec \theta$
• B. $\left(\sqrt{\frac{g}{l}}\right)\cos \theta$
• C. $\left(\sqrt{\frac{g}{l\cos \theta }}\right)\cos \theta$
• D. $\sqrt{\frac{g\sin \theta }{l}}$

Answer 1

The correct option is A $\left(\sqrt{\frac{g\sin \theta }{l}}\right)\sec \theta$

From wedge.
$cos\theta =\frac{BC}{l}$
$BC=lcos\theta$


From Free body diagram,
$N\sin \theta =m{\omega }^{2}R$
$N\cos \theta =mg$
Dividing (1) by (2),
$\frac{\sin \theta }{\cos \theta }=\frac{{\omega }^{2}R}{g}$
$\because R=BC=l\cos \theta$
${\omega }^2=\frac{g\sin \theta }{l{\cos }^2\theta }$
$\omega =\sec \theta \sqrt{\frac{g\sin \theta }{l}}$.