A man stands at the center of a platform that rotates without friction with an a

Q: A man stands at the center of a platform that rotates without friction with an a

a) Apply conservation of angular momentum. final angular momentum = initial angular momentum I2*w2 = I1*w1 w2 = I1*w1/I2 = 9.5*3.69/3.4 = 10.31 rev/s b) w1 = 3.69 revs = 3.69*2*pi rad/s = 23.18 rad/s w2 = 10.31 rev/s = 10.31*2*pi rad/s = 64.78 rad/s KI = 0.5*I1*w1^2 = 0.5*9.5*23.18^2 = 2552 J

ID: 1467120 • Letter: A

Question

A man stands at the center of a platform that rotates without friction with an angular speed of 3.69 rev/s. His arms are outstretched, and he holds a heavy weight in each hand. The moment of inertia of the man, the extended weights, and the platform is 9.5 kg*m^2. When the man pulls the weights inward toward his body, the moment of inertia decreases to 3.4 kg*m^2.

(a) What is the resulting angular speed of the platform?

(b) What is the kinetic energy of the system before the man pulls the weights to his body?

(d) What is the change in kinetic energy of the system?

Explanation / Answer

a) Apply conservation of angular momentum.

final angular momentum = initial angular momentum

I2*w2 = I1*w1

w2 = I1*w1/I2

= 9.5*3.69/3.4

= 10.31 rev/s

b) w1 = 3.69 revs = 3.69*2*pi rad/s = 23.18 rad/s

w2 = 10.31 rev/s = 10.31*2*pi rad/s = 64.78 rad/s

KI = 0.5*I1*w1^2

= 0.5*9.5*23.18^2

= 2552 J

Navigate

A man stands at the center of a platform that rotates without friction with an a

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

A man stands at the center of a platform that rotates without friction with an a

Question

Explanation / Answer

Related Questions

Navigate