The angular position of one of the arms of a spinning ice skater for 15 s...
The angular position of one of the arms of a spinning ice skater for 15 s is described by the function 1000 / (t + 5) rad for 0 ≤ t ≤ 15 where t is the elapsed time in seconds. The angular acceleration at t = 15 s is ____ rad / s².
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?
Problem 2. An ice skater is spinning at 60 RPM (counter-clockwise) when she pulls in her arms and undergoes an angular acceleration of 10 rad/s' for 0.6 seconds. What was her final RPM and how many rotations did she spin during this time?
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?
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?
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 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...
(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 0.470 kg-m 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 moment of inertia (in kg-m2) if his angular velocity drops to 1.35 rev/s. (c) Suppose instead he keeps his arms in and allows friction with the ice to slow him...
1. An ice skater spins on the ice with her arms positioned tight against her body. In this position, she has a moment of inertia of 1.3 kg m2 and an angular speed of 15 rad/s. If the ice skater then stretches out her arms, and her angular speed slows to 6.0 rad/s, what is her moment of inertia with her arms outstretched? 3.64 kg m2 4.91 kg m2 3.25 kg.m2 4.39 kg m2 6.11 kg m2 А В С...
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?