A 150-g piece of playdough slides across a frictionless table at v = 5.50 m/s. I

Question

A 150-g piece of playdough slides across a frictionless table at v = 5.50 m/s. It collides with a disk, I = jmr^2, of radius R = 35.0 cm and mass M = 2.50 kg which has a fixed frictionless axle. The playdough sticks to the disk. Treat the playdough as a point mass, I = mr^2. (a) If the disk is not rotating initially, what is its angular velocity after the collision? (b) What angular velocity would the disk need to have initially, if the disk stopped completely after the collision? L_0-playdough = L_playdough + L_disc rmv sintheta = I_bullet omega - I_door omega theta = sin^-1(20/35)

Explanation / Answer

(a) from conservation of angular momentum.

Ltotal before collision = Ltotal after collision

mvr =( Iplaydough + Idisc)*w

150g x 550 cm-s-1 x 20.0 cm = ( 150g x( 20.0 cm)2 + (1/2) x 2500g x (35.0)2)*w

1650000 s-1= ( 60000 + 1531250 )*w

w = 1650000/1591250 s-1

w = 1.04 s-1 ....................ans

(b) disk will stoped after collision if

initial angular momentum of disk = - angular momentum of playdough

hence

magnitude of  initial angular momentum of disk = magnitude of  initial angular momentum of playdough

Idisk*w = mvr

w = mvr/Idisk

= 1650000/1531250 s-1

= 1.08 s-1  ...................ans

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