Kinematics Question 490
Question: A long horizontal rod has a bead which can slide along it’s length, and initially placed at a distance L from one end A of the rod.The rod is set in angular motion about A with constant angular acceleration a. If the coefficient of friction between the rod and the bead it’s m, and gravity it’s neglected, then the time after which the bead starts slipping it’s[IIT-JEE Screening 2000]
Options:
A) $ \sqrt{\frac{\mu }{\alpha }} $
B) $ \frac{\mu }{\sqrt{\alpha }} $
C) $ \frac{1}{\sqrt{\mu \alpha }} $
D) Infinitesimal
Show Answer
Answer:
Correct Answer: A
Solution:
Let the bead starts slipping after time t For critical condition Frictional force provides the centripetal force $ m{{\omega }^{2}}L=\mu R=\mu m\times a _{t}=\mu Lm\alpha $
$
therefore $ m{{(\alpha t)}^{2}}L=\mu mL\alpha $
therefore $ t=\sqrt{\frac{\mu }{\alpha }} $ (As $ \omega =\alpha t $ )
