A student sits on a freely rotating stool holding two dumbbells, each of mass 3.

Question

A student sits on a freely rotating stool holding two dumbbells, each of mass 3.00 kg (see figure). When his arms are extended horizontally (Figure a), the dumbbells are 1.00 m from the axis of rotation and the student rotates with an angular speed of 0.750 rad/s. The moment of inertia of the student plus stool is 2.65 kg m^2 and is assumed to be constant. The student pulls the dumbbells inward horizontally to a position 0.250 m from the rotation axis (Figure b). Find the new angular speed of the student.

Explanation / Answer

Let’s first calculate the moments of inertia.
• First position:

Ii = I + 2mri^2
  
= 2.65 + 2 * 3 * 1^2

= 8.65 kg.m^2

• Second position

If = I + 2mrf^2

= 2.65 + 2 * 3 * 0.25^2

= 2.65 + 0.375

= 3.025 kg.m^2

Angular momentum is conserved:

Lf = Li

Ii * Wi = If * Wf

Wf = (Ii / If ) * w

= ( 8.65 / 3.025 ) * 0.750

= 2.14462809917 rad/s

Approx 2.14 rad/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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