1) An ice dancer with her arms stretched out starts into a spin with an angular

Question

1) An ice dancer with her arms stretched out starts into a spin with an angular velocity of 1.00 rad/s. Her moment of inertia with her arms stretched out is 2.48 kg m2. What is the increase in her rotational kinetic energy when she pulls in her arms to make her moment of inertia 1.40 kg m2?
A) 0.957 J
B) 0.902 J
C) 0.870 J
D) 0.750 J
E) 0.690 J

2) A 10 kg solid cylinder (I = 1/2 MR2) with a radius of 30.0 cm is rotating about a vertical axis through its center. If the angular momentum is increasing at the rate of 25 kg m2/s, then what is the torque?
A) 70 N · m
b) 45 N · m
C) 37 N · m
D) 25 N · m
E) 12 N · m

Explanation / Answer

there are no external torques acting, so her angular momentum stays constant, this means that

I before w before = I after w after

2,48 kgm^2 x 1 rad/s = 1.4 kgm^2 x w after

solve for w after = 1.77 rad/s

increase=2.913-1.24=0.870 J

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

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