**An ice skater is spinning at 5.2 rev/s and has a moment of inertia of 0.56…**

The following solution is suggested to handle the subject “**An ice skater is spinning at 5.2 rev/s and has a moment of inertia of 0.56…**“. Let’s keep an eye on the content below!

## Question “An ice skater is spinning at 5.2 rev/s and has a moment of inertia of 0.56…”

An ice skater is spinning at

5.2 rev/s and has a moment of inertia of 0.56 kg ⋅

m^{2}.

(9%) Problem 9: An ice skater is spinning at 5.2 rev/s and has a moment of inertia of 0.56 kg. m². > A 33% Part (a) Calculate the angular momentum, in kilogram meters squared per second, of the ice skater spinning at 5.2 rev/s. Grade Summary Deductions 0% Potential 100% L = 1 E sin() cos() tan() cotano asino acos atan acotan() sinh( cosh tanh cotanh0 Degrees Radians ( 7 8 9 HOME 4 5 6 * 1 2 3 – + – 0 . END VO BACKSPACE DEL CLEAR Submissions Attempts remaining: 5 (2% per attempt) detailed view Submit Hint Feedback I give up! Hints: 0% deduction per hint. Hints remaining: 1 Feedback: 0% deduction per feedback. A 33% 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. A 33% Part (c) Suppose instead he keeps his arms in and allows friction of the ice to slow him to 3.25 rev/s. What is the magnitude of the average torque that was exerted, in N.m, if this takes 19 s?

## Answer

L-IW I – moment inertio Wy Velocity X = 0.56X5.2X27 L = 16.29kg / m2 I.angular momentum conservation I. W = I2 W2 0.56×5.2 I 75 I = 3.63kg/m2 0. Forane – IX I = 0.56 kg 1 m2 x= Up – we a 0.564 (3025-5.2) 27 t torque torque = 0.36 Nin

## Conclusion

Above is the solution for “**An ice skater is spinning at 5.2 rev/s and has a moment of inertia of 0.56…**“. We hope that you find a good answer and gain the knowledge about this topic of **science**.