4. A 60.0-kg skater begins a spin with an angular speed of 7.5 rad/s. By changin

Question

4. A 60.0-kg skater begins a spin with an angular speed of 7.5 rad/s. By changing the position of her arms, the skater decreases her moment of inertia to one-half its initial value. a. What is the skater's angular momentum before she moves her arms? b. What is the skater's angular momentum after she moves her arms? c. What is the skater's final angular speed? A force is applied to the end of an 8.00 meter long uniform board weighing 300 N, the board also holds a Load of 150 N attached, in order to keep it horizontal, while it pushes against a wall at the left. If the angle the force makes with the board is 30° in the direction shown in figure then what is the magnitude of the applied force F? 5. Load

Explanation / Answer

4. I1 = I0

w1 = 7.5 rad/s

I2 = I0 / 2

(A) suppose moment of inertia initially was I0 then

L = I0 w = 7.5 I0

(b) angular momentum will be conserved.

Lf = 7.5 I0

(c) 7.5 I0 = I2 w2

7.5 I0 = (I0/2) w2

w2 = 15 rad/s ..........Ans

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.