In the figure, three 0.03 kg particles have been glued to a rod of length L 1 1

Question

In the figure, three 0.03 kg particles have been glued to a rod of length L 1 1 cm and negligible mass and can rotate around a perpendicular axis through point O at one end. How much work is required to change the rotational rate (a) from 0 to 20.0 rad/s, (b) from 20.0 rad/s to 40.0 rad/s, and (c) from 40.0 rad/s to 60.0 rad/s? (d) What is the slope of a plot of the assembly's kinetic energy (in joules) versus the square of its rotation rate (in radians-squared per second-squared)? Axis (a) Number (b) Number (c) Number (d) Number Units ( Units Units Units

Explanation / Answer

a)

L = 3 d

d = L/3 = 11/3 = 0.11/3 = 0.0367

I = moment of inertia of the system = m (d2 + (2d)2 + (3d)2) = (14) m d2 = (14) (0.03) (0.0367)2 = 5.66 x 10-4 kgm2

wi = initial angular velocity = 0 rad/s

wf = final angular velocity = 20 rad/s

W = work done = change in rotational KE = (0.5) I (wf2 - wi2) = (0.5) (5.66 x 10-4) ((20)2 - (0)2) = 0.113 J

b)

wi = initial angular velocity = 20 rad/s

wf = final angular velocity = 40 rad/s

W = work done = change in rotational KE = (0.5) I (wf2 - wi2) = (0.5) (5.66 x 10-4) ((40)2 - (20)2) = 0.34 J

c)

KE = (0.5) I w2

so KE/w2 = (0.5) I

it represent half of the moment of inertia

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