A 60-kg skater that spins 2 complete revolutions in 1.0 seconds while in the air has...
A skater has a moment of inertia of 100 kg . m^2 when his arms are outstretched and a moment of inertia of 75 kg . m^2 when his arms are tucked in dose to his chest. If he starts to spin at an angular speed of 2.0 rps (revolutions per second) with his arms outstretched, what will his angular speed be when they are tucked in?
Dirk the ice skater spins at 4.51 rev/s and has moment of inertia is 0.56 kg ⋅ m2 . If he decreases his rate of spin to 2.45 rev/s by spreading his arms, what is his new 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...
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...
Problem 2: An ice-skater, as we mentioned in lecture, in order to increase her angular velocity from 2.0 rev per 1.3 sec to 3.5 rev per sec she needs to decrease her moment of inertia to a value of 4.6 kg m/sec by pulling hers arms towards her body. a) Find her initial moment of inertia when her arms are out-stretched. b) Calculate the rotational kinetic energy for each case.
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...
I don't understand how to find 6b and all of 7. 6. A hollow sphere with mass = 0.65 kg and radius = 0.13 m is initially at rest on a 20° incline and rolls down the incline without slipping. The initial height of the disk (H) a. At the top of the incline, just before the disk begins to roll, what is the total mechanical energy of the disk? Emech=PEtop=6.37) b. Determine the velocity of the disk at H=...
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
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)?
Problem 19: An ice skater is spinning at 6.2 rev/s and has a moment of inertia of 0.36 kg ⋅ m2.Part (a) Calculate the angular momentum, in kilogram meters squared per second, of the ice skater spinning at 6.2 rev/s. Part (b) 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 0.75 rev/s. Part (c) Suppose instead he keeps his...