A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It

Question

A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It is attached to a massless cord passing through a hole in the surface. The block is originally revolving at a distance 0.50 m from the hole with a speed of 0.855 m/s. The chord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.25 m. a distance 0.50 m from the hole with a speed of 0.855 Treat the block as a particle. (a) Is the angular momentum of the block conserved, why or why not? (b) What is the angular speed at the new radius? (c) What is the tension in the cord? (c) Find the change in the kinetic energy of the block. (d) How much energy was done in pulling the cord? (e) Suppose that the block is constrained at this radius and the tension removed, how long will t take it to stop if it is subject to a braking acceleration of 0.5 rad/s? (f) How many revolutions it will have done before stopping?

Explanation / Answer

(a) Net torque due to tension is zero.

hence tension will not change the angular momentum .

so angular momentum is conserved.

(b) m v1 r1 = m v2 r2

0.50 x 0.855 = v2 x 0.25

v2 = 1.71 m/s

angular speed = 1.71 / 0.25 = 6.84 rad/s

(C) T = m w^2 r =0.0250 x 6.84^2 x 0.25

T = 0.29 N

(c) KEi = m vi^2 /2 = 0.0250 x 0.855^2 / 2

= 9.138 x 10^-3 J

KEf = 0.0250 x 1.71^2 /2 = 36.6 x 10^-3 J

change in KE = KEf - KEi = 27.5 x 10^-3 J

(D) Energy done = 27.5 x 10^-3 J

(E) wf = wi + alpha t

0 = 6.84 - 0.5 t

t = 13.68 sec

(F) wf^2 - wi^2 =2 alpha theta

0^2 - 6.84^2 = 2(-0.5) theta

theta = 46.8 rad

revolutions = 46.8 / 2pi = 7.5 revolutions

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