Two skaters, each of mass 60 kg, approach each other along parallel paths separa

Question

Two skaters, each of mass 60 kg, approach each other along parallel paths separated by 12.1 m. They have equal and opposite velocities of 1.2 m/s. The first skater carries one end of a long pole with negligible mass, and the second skater grabs the other end of it as she passes; see the figure. Assume frictionless ice. Describe quantitatively the motion of the skaters after they have become connected by the pole. What is their angular speed? By pulling on the pole, the skaters reduce their separation to 1.0 m. What is their angular speed then? Calculate the ratio of the final kinetic energy to the original kinetic energy

Explanation / Answer

a) w = v/r

here, r = L/2 = 4.9/2 = 2.45 m

so, w = 1.9/2.45

= 0.78 rad/s <<<<<<<<<<<<<<<<<<-------------Answer

b) Apply consrvation angular momentum

I1*w1 = I2*w2

w2 = w1**I1/I2

= w1*(2*m*r1^2)/(2*m*r2^2)

= w1*(r1/r2)^2

= 0.78*(2.45/0.5)^2

= 18.7 rad/s <<<<<<<<<<<<<<<<<<-------------Answer

c)

KEf = 0.5*I2*w2^2

= 0.5*(2*m*r2^2)w2^2

KEi = 0.5*(2*m*r1^2)w1^2

KEf/KEi = (r2^2*w2^2)/(r1^2*w1^2)

= (0.5^2*18.7^2)/(2.45^2*0.78^2)

= 24.2 <<<<<<<<<<<<<<<<<<-------------Answer

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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