1. An ice skater spins on the ice with her arms positioned tight against her body....
An ice skater spins, with her arms and one leg outstretched, and achieves an angular velocity of 2 rad/s. when she pulls in her arms, her moment of inertia decreases to 65% its original value. what is her new angular velocity?
Please write legibly and write out all equations and units. Please show all steps. Algebra-based physics only please. 5. An ice skater spins on the ice with her arms tight against her body. In this position, she has a moment of inertia of 1.2 kg.m² and an angular speed of 15 rad/s. She then stretches her arms out and her angular speed slows to 6.0 rad/s. What is her new moment of inertia? A B C D E 4.2 kg.m...
An ice skater spinning with outstretched arms has an angular speed of 5.0rad/s . She tucks in her arms, decreasing her moment of inertia by 29% . What is the resulting angular speed? rad/s By what factor does the skater's kinetic energy change? (Neglect any frictional effects.) where does the extra kinetic energy come from?
An ice skater has a moment of inertia of 5.0 kg-m2 when her arms are outstretched. At this time she is spinning at 3.0 revolutions per second (rps). If she pulls in her arms and decreases her moment of inertia to 2.0 kg-m2, how fast will she be spinning? A) 7.5 rps B) 8.4 rps C) 2.0 rps D) 10 rps E) 3.3 rps
A 50 kg ice skater spins about a vertical axis through her body with her arms horizontally outstretched, making 2.5 turns each second. The distance from one hand to the other is 1.50 m. Biometric measurements indicate that each hand typically makes up about 1.25 % of body weight. What horizontal force must her wrist exert on her hand? Express the force in part (a) as a multiple of the weight of her hand.
Assume an ice skater in the ending position, with arms and legs folded in, has a moment of inertia of 0.80 kg*m2. Also assume the skater starts with both arms and one leg out and has a moment of inertia in this configuration of 3.2 kg*m2. If he ends spinning at 1.3 rev/s, what is his angular speed (in rev/s) at the start?
Natalie is an accomplished ice skater with hopes of competing in the 2022 Winter Olympics in Beijing. One of her standard moves is to spin on point. She starts spinning at 3.5 rev/s with her arms outstretched and an associated moment of inertia I = 6.4 kg ∙ m2. Natalie then brings her arms in and decreases her moment of inertia to I = 1.8 kg ∙ m2. What is her final angular speed? A. 10 rev/s B. 3.5 rev/s...
A skater has a moment of inertia of 4kg.m2 when both her arms are outstretched rotating at 60 rpm. When she draws her arms in her moment of inertia drops to 0.8kg.m2 . What is her angular momentum and new speed of rotation in rpm?
(a) Calculate the angular momentum (in kg.m2/s) of an ice skater spinning at 6.00 rev/s given his moment of inertia is 0.470 kg.m2 kg-m2/s (b) He reduces his rate of spin (his angular velocity) by extending his arms and increasing his moment of inertia. Find the value of his moment of inertia (in kg m-) if his angular velocity drops to 1.00 rev/s. kg-m2 (c) suppose instead he keeps his arms in and allows friction with the ice to slow...
An ice skater with moment of inertia 70.0 kg•m2 is spinning at 41.0 rpm. If the skater pulls in her arms, her moment of inertia decreases to 50.0 kg•m2. What is the skater’s resulting angular velocity?