A skater rotates at 3 revolutions per second with her arms stretched. What must she do to decrease her moment of inertia?
Need more information
stretch both arms and foot out.
stretch her foot out.
pull her arms in
pull her arm in
will reduce here the moment of inertia
because I = m*r2
so you see the moment of inertia depends on distance so if she bring her arms closer to axis moment of inertia decreases
A skater rotates at 3 revolutions per second with her arms stretched.
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?
A figure skater is spinning at a rate of 0.80 revolutions per second with her arms close to her chest. She then extends her arms outwards and her new rotational frequency is 0.40 revolutions per second. What is ratio of her new moment of inertia to her original moment of inertia?
a figure skater us soinning at a rate if .8 revolutions per second witg her arms close to her chest. She rgen extends her arms outwRd and her new rotational frequency is .4 revolutions per second. what raio of her new moment of inertia to her oroginal moment of inertia?
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
An ice skater is spinning at 2.5 revolutions per second and has a moment of inertia of 0.85 kg m2. Estimate her rotational angular momentum, assuming for simplicity that she can be approximated as a rigid, axially-symmetric body.
A 50.0kg mass skater with her arms stretched out giving her a radius of 0.9m is spinning with an angular velocity of 15 rad/s. If she pulls her arms close to her body her new radius is 0.30m, what is her new angular velocity?
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.
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?
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...
4 A ballerina is spinning on one foot with her arms stretched out, when she pulls her arms in. Will the following quantities increase, decrease or remain unchanged? (Ignore air drag) (i) Her angular speed. Increase (ii) Her moment of inertia. Decrease (iii) Her angular momentum. Increase (iii) Her rotational kinetic energy. Increase. K1 = 1/2 I1ω1 ; K2 = 1/2 I2ω2 . Because I1ω1 = I2ω2 , it follows K2 = (1/2 I1ω1)(I1/I2) = K1(I1/I2) > K1 THE ANSWERS...