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A long horizontal rod has a bead which can slide along its length and initially placed at a distance L form one end A of the rod. The rod is set in angular motion about A with constant angular acceleration . If the coefficient of friction between the rod and the bead is and gravity is neglected, then the time after which the bead start slipping is

A long horizontal rod has a bead which can slide along its length and initially placed at a distance L form one end A of the rod. The rod is set in angular motion about A with constant angular acceleration . If the coefficient of friction between the rod and the bead is  and gravity is neglected, then the time after which the bead start slipping is


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Grade:11

2 Answers

Arun
25763 Points
10 months ago
Assuming the given motion takes place in Horizontal plane.
 
Steps:
1) The bead will start sliding when centripetal force is equal to limiting friction.
i. e.
mw^2L = f(max)
=> mw^2 L= u N ----(1)
 
2) We know that,
Here. Normal Reaction is provided by Tangential force.
F(t) = mLα =N ---(2)
 
3 ) Substituting value of N in eq. (1),
Now,
At time t,.
w = 0 + αt
w = αt
 
Now,
mw^2L = umLα
=> m(αt) ^2 L = umLα
=> αt^2 = u
=> t =√(u/α) .
 
At time t , bead will start sliding.
Vikas TU
14149 Points
9 months ago
When we are giving an angular acceleration to the rod, the bead is also having an instantaneous. acceleration a=La. This will happen when a force is exerted on the bead by the rod. The bead has a tendency to move away from the centre. But due to the friction between the bead and the rod, this does not happen to the extent to which frictional force is capable of holding the bead. The frictional force here provides the necessary centripetal force. If instantaneous angular velocity is o then 
 
mw^2L = umLα
=> m(αt) ^2 L = umLα
=> αt^2 = u
=> t =√(u/α) .
 
At time t , bead will start sliding

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