A small block on a frictionless, horizontal surface has a mass of 270times10^-2k

Question

A small block on a frictionless, horizontal surface has a mass of 270times10^-2kg. It is attached to a massless cord passing through a hole in the surface (Figure 1). The block is originally revolving at a distance of 0.300 m from the hole with an angular speed of 2.61 rad/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 m. Model the block as a particle. Is angular momentum of the block conserved? What is the new angular speed? Find the change in kinetic energy of the block. How much work was done in pulling the cord? Express your answer with the appropriate units.

Explanation / Answer

Part A:

Yes. Angular momentum of the block shall always be conserved.

Part B:

Angular momentum of block = moment of inertia (I) x angular velocity ()
I = mR² (point mass)
Applying cons. of ang. momentum .. Ii.i = If.f

(m.0.30²) x 2.61 = (m.0.15²) x f

=> f = 0.3² x 2.61/ 0.15² = 10.44 rad/sec.

Part C:
Angular KE = ½ I²

KEi = ½ x (m.0.3²) x 2.61²
KEf = ½ x (m.0.15²) x 10.44²

KE = KEf - KEi
KE = ½m {(0.15² x 10.44²) - ( 0.3² x 2.61²)} =  ½m {2.452 - 0.613}

KE = ½ x 0.027kg {1.839} =>KE = 24.83x 10^-3 J This is gain in kinetic energy.

Part D:

WD (J) = gain in KE (J) =>WD = 24.83x 10^-3 J

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