Circular Motion 5) A child sits on a merry-go-round, 1.5 meters from the center.

Question

Circular Motion 5) A child sits on a merry-go-round, 1.5 meters from the center. The merry-go-round is turning at a constant rate, and the child is observed to have a radial acceleration of 2.3 m/s. How long does it take for the merry- go-round to make one revolution? 6) Two particles, A and B, are in uniform circular motion about a common center. The acceleration of particle A is 4.7 times that of particle B. Particle B takes 2.4 times as long for a rotation as particle A. The ratio of the radius of the motion of particle A to that of particle B is closest to: 7) A wind farm generator uses a two-bladed propeller mounted on a pylon at a height of 20 m. The length of each propeller blade is 12 m. A tip of the propeller breaks off when the propeller is vertical. The fragment flies off horizontally, falls, and strikes the ground at P. Just before the fragment broke off, the propeller was turning uniformly, taking 1.2 s for each rotation. a) What is the distance from the base of the pylon to the point where the fragment strikes the ground? b) What is the angle with respect to the vertical at which the fragment strikes the ground?

Explanation / Answer

5)

v^2 = 1.5 x 2.3 = 3.45

v = 1.8574 m/s

then t = 2pi*1.5/v = 2pi*1.5/ 1.8574 = 5.074 sec

6)

a[A] = r[A]w[A]^2
a[B] = r[B]w[B]^2
where w[A] and w[B] are the angular velocities of A and B.
a[A] = 4.7*a[B]

T[A] = 2.pi/w[A]
T[B] = 2.pi/w[B] = 2.4*T[A]
T[B]/T[A] = 2.4 = w[A]/w[B] ==> w[A]^2/w[B]^2 = 5.76

a[A]/a[B] = 4.7 = r[A]w[A]^2/r[B]w[B]^2 = 5.76*(r[A]/r[B])
r[A]/r[B] = 4.7/5.76 = 0.816

7)

The vertical velocity depends only on the height

The time to hit the ground comes from y = 1/2*g*t^2

so t = sqrt(2*y/g) = sqrt(2*32m/9.8) = 2.555s

Therefore the vertical velocity = g*t = 9.8m/s^2*2.555s = 25.04m/s

Now the horizontal velocity comes from the rotational motion

We have the angular velocity 2pi/t = 2pi/1.2s = 5.26rad/s

So the linear velocity at the top = r*5.26 = 12*5.26 = 62.8m/s

So the angle with respect to the vertical = arctan(vx/vy) = arctan(62.8/25.04) = 68o

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.