5. After rotating at the angular speed of 2.5 radians/second, the merry-go-round

Question

5. After rotating at the angular speed of 2.5 radians/second, the merry-go-round slows down with a constant angular acceleration stopping after 4 complete revolutions. What is the magnitude of the angular acceleration of the merry-go-round? (a) 0.124 radians/s (b) 0.750 radians/s (c) 1.23 radians/s2 A triangular shape is made from identical balls and identical rigid, massless rods as shown. The moment of inertia about the a, b, and c axes isI.I, and I respectively Calculate la, Ib and Ic 6. Find la 8mr2 7. Find Ib- 3mr2 8. Find lc= 4 9. Which of the following orderings is correct?

Explanation / Answer

First Question -

(5) Option (a) is the correct answer.

Explanation -

Initial angular speed, w0 = 2.5 rad/sec.

merry-go-round slows down after completing 4 complete revolutions.

so, theta = 4 x 2 x pi radian = 25.13 rad.

suppose, alfa = angular acceleration

use the expression -

w^2 = w0^2 + 2*alfa*theta

=> 0 = 2.5^2 + 2*alfa*25.13

=> alfa = -2.5^2 / (2*25.13) = -0.124 rad/s^2

so the magnitude is 0.124 rad/s^2

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