Question

A rod is given an angular acceleration α= 3 rad/s^2 from rest so that it rotates in a horizontal plane about a vertical axis. It has a ring at a distance r from the axis of rotation. The friction coefficient between the ring and the rod is µ= 1/3. Neglecting gravity find the time after which the ring will start to slip on the rod.

Answer 1

Answer:

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Explanation:

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Answer 2

Concept

Rotational acceleration is another name for angular acceleration. It expresses the change in angular velocity per unit time quantitatively.

Given

Given that Angular acceleration = 3rad/s^2

Friction coefficient = µ= 1/3

Find

We need to find the time  after which the ring will start to slip on the rod.

Solution

We know that

ω = αt

Let the ring flips at time t

Then force applied on the ring is

mω²r = µmg

∴ ω² = µg/r

∴α²t² =  µg/r

∴  t = $\sqrt{\frac{ug}{r}*\frac{1}{\alpha } }$

∴ t = $\frac{1}{3\sqrt{3}} \sqrt{\frac{g}{r} }$

Hence Time is $\frac{1}{3\sqrt{3}} \sqrt{\frac{g}{r} }$ seconds

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