A playground ride consists of a disk of mass M = 40 kg and radius R = 2.4 m moun

Question

A playground ride consists of a disk of mass M = 40 kg and radius R = 2.4 m mounted on a low-friction axle. A child of mass m = 24 kg runs at speed v = 2.4 m/s on a line tangential to the disk and jumps onto the outer edge of the disk. (d) If the disk was initially at rest, now how fast is it rotating? That is, what is its angular speed? (The moment of inertia of a uniform disk is A1/2MR^2.) omega = radians/s (g) What was the speed of the child just after the collision? v - m/s (j) Calculate the change in linear momentum of the system consisting of the child plus the disk (but not inducing the axle), from just before to just after impact, due to the impulse applied by the axle. Take the x axis to be in the direction of the initial velocity of the child. Delta p = p_x, f - P_x, i = kg A middot m/s (k) The child on the disk walks inward on the disk and ends up standing at a new location a distance R/2 - 1.2 m from the axle. Now what is the angular speed? (It helps to do this analysis algebraically and plug in numbers at the end.) omega = radians/s

Explanation / Answer

The angular momentum is conserved as there is no angular impulse:

d) initial angular momentum = m v R = 24*2.4*2.4 = 138.24 kg.m^/s^2

final momentum = omega*(MR^2/2+mR^2) = omega*(40*2.4^2/2+24*2.4^2) = 253.44 omega

So 253.44 omega = 138.24

omega = 138.24/253.44 = 0.545 rad/s

g) Speed after collision = omega*R = 0.545*2.4 = 1.308 m/s

j) The change in linear momentum of the disk is zero.

Hence total change in linear momentum due to axle impulse = m(v_f-v_i) = 24*(1.308-2.4) = -26.208 kg m/s

k)

I_1*omega_1 = I_2*omega_2

(MR^2/2+mR^2)*omega_1 = (MR^2/2+mR^2/4)*omega_2

(40*2.4^2/2+24*2.4^2)*0.545 = (40*2.4^2/2+24*2.4^2/4)*omega_2

omega_2 = 0.922 rad/s.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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