A small block on a frictionless surface has a mass of 90 g. It is attached to a

Question

A small block on a frictionless surface has a mass of 90 g. It is attached to a massless string passing through a hole in a horizontal surface (see diagram). The block is originally rotating in a circle of radius 65 cm with angular speed 0.70 rad/s. The string is then pulled from below until the radius of the circle is 45 cm. You may treat the block as a point particle.

(a) Is the angular momentum of the block conserved? Why or why not? (b) What is the final angular speed?
(c) What are the initial and final tensions in the string? (d) What was the change in kinetic energy of the block?

(e) How much work was done in pulling the string?

Explanation / Answer

yes . angular momentum remain conserved .

there is no external torque being applied to change it.

part b )

Li = Lf

I*w^2 = Iw^2

I for point = mr^2

r1^2*w1^2 = r2^2 * w2^2

w2^2 = r1^2 * w1^2 / r2^2

w2^2 = 0.65^2 * 0.70^2 / 0.45^2

w2 = 1.01 rad/s

part c )

Tension is equal to centripetal force .

intial

T = mw^2*r

T = 90*10^-3 * (.70)^2 * 0.65

T = 0.028665 N

tension final

T = mw^2*r

Tf = 0.041405 N

part d)

KE = 1/2*Iw^2

change in KE = 1/2*m(w2^2 * r2^2 - w1^2 * r1^2)

dKE = -4.5475 x 10^-4 J

part e )

change in KE = work done

w = -4.5475 x 10^-4 J

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