Find the time at which the bead will start slipping

[Kaushik]

Homework Statement

A long horizontal rod has a bead which can slide along its length and is initially placed at a distance $L$ from one end A of the rod.The rod starts from rest in angular motion about A with a constant angular acceleration $\alpha$ .If the coefficient of friction between the rod and the bead is $µ$ and gravity is neglected then what is the time after which the bead starts slipping.

Relevant Equations

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A long horizontal rod has a bead which can slide along its length and is initially placed at a distance $L$ from one end A of the rod.The rod starts from rest in angular motion about A with a constant angular acceleration $\alpha$ .If the coefficient of friction between the rod and the bead is $µ$ and gravity is neglected then what is the time after which the bead starts slipping.

 

[PeroK]

Problem Statement: A long horizontal rod has a bead which can slide along its length and is initially placed at a distance $L$ from one end A of the rod.The rod starts from rest in angular motion about A with a constant angular acceleration $\alpha$ .If the coefficient of friction between the rod and the bead is $µ$ and gravity is neglected then what is the time after which the bead starts slipping.
Relevant Equations: -

A long horizontal rod has a bead which can slide along its length and is initially placed at a distance $L$ from one end A of the rod.The rod starts from rest in angular motion about A with a constant angular acceleration $\alpha$ .If the coefficient of friction between the rod and the bead is $µ$ and gravity is neglected then what is the time after which the bead starts slipping.

How far can you get on your own?

 

[Kaushik]

How far can you get on your own?

• Centripetal force is the frictional force.
• With respect to the bead, it experiences a frictional force which pulls it towards the center with radius = L and centrifugal force (pseudo force). $ƒ_s ≤ µN = m\omega^2L$
• $\alpha = \omega /t=> \omega^2 = (\alpha t)^2$
• Normal force acting on it is given the tangential acceleration, $N =(m)a_t = (m)\alpha L$
• $µ(m)(\alpha L) ≤ m \omega ^2 L = m(\alpha)^2(t)^2(L)$
• $t ≥ \sqrt{\frac{µ}{\alpha}}$

Is it correct?

 

[PeroK]

• Centripetal force is the frictional force.
• With respect to the bead, it experiences a frictional force which pulls it towards the center with radius = L and centrifugal force (pseudo force). $µN = m\omega^2L$
• $\alpha = \omega /t=> \omega^2 = (\alpha t)^2$
• Normal force acting on it is given the tangential acceleration, $N =(m)a_t = (m)\alpha L$
• $µ(m)(\alpha L) = m \omega ^2 L = m(\alpha)^2(t)^2(L)$
• $t = \sqrt{\frac{µ}{\alpha}}$

Is it correct?

Sorry, I'm not getting alerts any more, so I missed your post.

Yes, that all looks good. One minor point is that you are dealing with the maximum possible frictional force. The actual frictional force is less than this, and is equal to the required centripetal force.

 

[Kaushik]

Yes, that all looks good. One minor point is that you are dealing with the maximum possible frictional force. The actual frictional force is less than this, and is equal to the required centripetal force.

Is the edited answer correct?

 

[PeroK]

Is the edited answer correct?

When I looked at this, I wrote:

Let $f_s = \mu N$ be the maximum frictional force.

That avoids any confusion (or potential loss of marks in an exam). If you simply use a frictional force of $\mu N$ without any comment, then you risk a pedantic examiner taking issue with you.

In general it's a good idea always to describe $\mu N$ as the maximum frictional force. That was all I meant. Your solution stood as it was. But, you didn't say precisely what $f_s$ denoted.