lecture_13 - Circular motion and unbalanced force A...

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PH1011 Mechanics Circular motion and unbalanced force A particle moving with speed v around the circumference of a circle of radius r has an acceleration of v 2 / r directed towards the centre. From N2, acceleration implies action of an unbalanced force . Example : a stone whirled around in a horizontal circle on the end of a string. The tension in the string is the force that supplies the acceleration. In problems like this involving circular motion, we need to: identify the unbalanced force, then apply N2 to equate it to (mass x acceleration), i.e. mv 2 / r . The resulting equation provides a condition that must be satisfied in order to maintain the circular motion that was originally assumed.
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PH1011 Mechanics Example Problem : A string 1m in length can just sustain a force of 200 N without breaking. A particle of mass 2 kg is attached to one end and revolves uniformly on a smooth table, the other end being fixed. Find the maximum rotational velocity of the particle. Solution : Maximum allowable ω=ω max gives tension of 200 N in string. This force must be equal to m ω 2 r by N2. So from which we find ω max = 10 rad s -1 . 200 = m ω max 2 r = 2. max 2 .1
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PH1011 Mechanics Centripetal force and acceleration: some additional remarks Centripetal force is a very real force (see following examples), but it has one major difference to all the forces we have studied so
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This note was uploaded on 11/23/2010 for the course CIVIL 105 taught by Professor No during the Spring '10 term at Lovely Professional University.

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lecture_13 - Circular motion and unbalanced force A...

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