Question

A merry-go-round rotates from rest with an angular acceleration of 1.08 rad/s2. How long does it take to rotate through (a) the first 3.74 rev and (b) the next 3.74 rev

Answer 1

Let ω1 be the initial angular speed and ω2 the final angular speed:

α = (ω2- α1) / t

corresponding to a = (v2 - v1) / t

S (distance corresponds to theta)

1 rev = 2 pi and 3.74 rev = 7.48 pi = 23.5 radians

S = 1/2 a t^2 linear or S = 1/2 α t^2 angular acceleration

23.5 = 1/2 * 1.08 t^2 and t = 6.60 sec for first 3.84 rev

b) ω1 = 1.08 * 6.6 = 7.13 rad/sec initial speed for second 3.74

23.5 = 7.13 t + .54 t^2 compare to S = v1 t + 1/2 a t^2

.54 t^2 + 7.13 t - 23.5 = 0

t^2 + 13.2 t - 43.5 = 0

t = 2.7 sec for next 3.74

Check:

7.13 * 2.7 + .54 * 2.7^2 = 23.2 rad = 3.7 rad

Answer 2

The time taken to rotate through the first 3.74 rev and the next 3.74 rev is 10.33 seconds.

To find the time taken to rotate through a certain number of revolutions, we can use the formula:

t = √(2πn/α)

where t is the time, n is the number of revolutions, and α is the angular acceleration.

(a) For the first 3.74 revolutions:

t = √(2π(3.74)/1.08) = 10.33 seconds

(b) For the next 3.74 revolutions:

t = √(2π(3.74)/1.08) = 10.33 seconds

Therefore, The time taken to rotate through the first 3.74 rev and the next 3.74 rev is 10.33 seconds.