a parcticle attached to a string of length 2 m is given an initial velocity of 6 m/s the string is attachhed to a peg and as the particle rotates about the peg the string winds around the peg.what lenght of string has wound around the peg when the velocity of the particle is 20 m/s?

Asked by Rahul SAHANI | 19th Jan, 2014, 08:38: PM

Expert Answer:

As the string winds around the peg, the radius of rotation of the particle decreases, causing a decrease in the moment of inertia of the particle.

Here momentum is conserved and, as the moment of inertia of the particle decreases, its speed increases.( Recall that v = ωr).

Thus the initial angular velocity of the particle,

ω0 = v/r = 3 rad/s

The initial moment of inertia of the particle is,

 I0 = mR 2 = 4m 

 We want to find r , the radius of the string when the particle has a speed of 20 m/s.

At this point, the angular velocity of the particle is,

ωf = v/r = 20/r

The moment of inertia =  I= mr2 

Now apply the conservation of angular momentum to find our value for r.

Initial angular momentum = final angular momentum

I0ω0 = Ifωf

(4m) × 3 = mr2 × 20/r

12 = 20 r

r = 0.6 m



Answered by Komal Parmar | 20th Jan, 2014, 12:36: AM

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