A playground ride consists of a disk of mass M = 50 kg and radius R = 2.4 m moun

Question

A playground ride consists of a disk of mass M = 50 kg and radius R = 2.4 m mounted on a low-friction axle. A child of mass m = 16 kg runs at speed v = 2.8 m/s on a line tangential to the disk and jumps onto the outer edge of the disk. ANGULAR MOMENTUM (a) Consider the system consisting of the child and the disk, but not including the axle. Which of the following statements are true, from just before to just after the collision? The axle exerts a force on the system but nearly zero torque. The torque exerted by the axle is nearly zero even though the force is large, because || is nearly zero. The angular momentum of the system about the axle changes. The momentum of the system doesn't change. The momentum of the system changes. The angular momentum of the system about the axle hardly changes. The torque exerted by the axle is zero because the force exerted by the axle is very small. (b) Relative to the axle, what was the magnitude of the angular momentum of the child before the collision? |C| = kg·m2/s (c) Relative to the axle, what was the angular momentum of the system of child plus disk just after the collision? |C| = kg·m2/s (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 ½MR2.) = radians/s (e) How long does it take for the disk to go around once? Time to go around ENERGY (f) If you were to do a lot of algebra to calculate the kinetic energies before and after the collision, you would find that the total kinetic energy just after the collision is less than the total kinetic energy just before the collision. Where has most of this energy gone? Increased translational kinetic energy of the disk. Increased thermal energy of the disk and child. Increased chemical energy in the child.

Explanation / Answer

A) here the angular momentum is conserved

so the answers are

the torque ecerted by the axle is nearly zero even though the force is large,because is zero

the momentum of the system changes

b) angular momentum of the child before collision is Li = m*v*r = 16*2.8*2.4 = 107.54 kg-m^2/sec

c) according to law of conservation of angular momentum

angular momentum of the child + disk after collision = 107.54 kg-m^2/sec

d) angular momentum of the system before collision = angular momentum of the system after the collision

(m*v*r)+I1*w1 = I2*w2

since w1 = 0 rad/s

107.54 + 0 = ((0.5*M*R^2)+(m*R^2))*w2

107.54 = ((0.5*50*2.4^2)+(16*2.4^2))*w2

w2 = 0.46 rad/s

e)using kinematic equations

w2^2-w1^2 = 2*alpha*theta

for one complete round angular displacement is theta = 2*pi rad

0.46^2 - 0^2 = 2*alpha*2*3.142

angular accelaration is alpha = 0.46^2/(2*2*3.142)

alpha = 0.0168 rad/s^2

but alpha = (w2-w1)/t

t = (w2-w1)/alpha = (0.46)/0.0168 = 27.4 sec

f) increased thermal energy of the disk and child

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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