A figure skater is spinning on frictionless ice. Treat the figure skater as a sphere with...
An ice- skater is initially spinning at an angular speed ω = 1.35 revolutions/s with a rotational inertia Ii = 2.30 kg.m2 with her arms extended. When she pulls her arms in, her rotational inertia is reduced to If=1.05 kg.m2 . Assume no external torques act. a) Determine her initial angular speed in rad/s. (1 marks) b) Calculate her final angular speed in RPM (4 marks) c) Calculate the period of rotation when she is at her final speed (1...
4. An ice skater with rotational inertia I = 0.23 kg*m* is spinning with angular speed w. They pull their arms in, increasing their angular speed to 4w. What is the final moment of inertia?
An ice- skater is initially spinning at an angular speed ω = 1.35 revolutions/s with a rotational inertia Ii = 2.30 kg.m2 with her arms extended. When she pulls her arms in, her rotational inertia is reduced to If=1.05 kg.m2 . Assume no external torques act. a) Determine her initial angular speed in rad/s. (1 marks) b) Calculate her final angular speed in RPM (4 marks) c) Calculate the period of rotation when she is at her final speed (1...
3. An ice skater starts spinning at a rate of 2.0 rev/s with their arms extended. They then pull their arms in toward their body reducing their moment of inertia by ¼, what is the angular velocity of the skater with their arms pulled in?
If an ice skater has a rotational inertia of 100 kg*m^(2)while spinning with an angular velocity of 2 rad/s, what is the ice skaters angular velocity if she changes her rotational inertia to 50 kg*m^(2)?
An ice skater is spinning at 6.8 rev/s and has a moment of inertia of 0.24 kg ⋅ m2. Calculate the angular momentum, in kilogram meters squared per second, of the ice skater spinning at 6.8 rev/s. He reduces his rate of rotation by extending his arms and increasing his moment of inertia. Find the value of his moment of inertia (in kilogram meters squared) if his rate of rotation decreases to 1.25 rev/s. Suppose instead he keeps his arms...
Calculate the angular momentum, in kg · m2/s, of an ice skater
spinning at 6.00 rev/s given his moment of inertia is 0.330 kg ·
m2.
(a) Calculate the angular momentum, in kg . m/s, of an ice skater spinning at 6.00 rev/s given his moment of inertia is 0.330 kg . m2. kg. m/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...
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?
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?
A figure skater is spinning at a rate of 0.75 revolutions per second with her arms close to her chest. She then extends her arms outwards and her new rotational frequency is 0.50 revolutions per second. What is ratio of her new moment of inertia to her original moment of inertia?