A man stands on a platform that is rotating (without friction) with an angular s

Question

A man stands on a platform that is rotating (without friction) with an angular speed of 1.27 rev/s; his arms are outstretched and he holds a brick in each hand. The rotational inertia of the system consisting of the man, bricks, and platform about the central axis is 5.91 kg·m2. If by moving the bricks the man decreases the rotational inertia of the system to 1.89 kg·m2, (a) what is the resulting angular speed of the platform and (b) what is the ratio of the new kinetic energy of the system to the original kinetic energy?

A man stands on a plattorm that is otating without friction with an angular speed o 1 27 ev s his arms are outstretched and he holds a brick in each hand. The rotational inertia o the system consisting of the man, bricks, and platform about the centra axis is 5.91 If by moving the bricks the man decreases the rotational Inertla of the system to 1.89 kgm2, (a) what ls the resulting angular speed of the platform and (b) what is the ratlo of the new kinetic energy of the system to the original kinetic energy (a) Number (b) Number 2 ew ldneticenconsisting of Units Units

Explanation / Answer

a)

Conservation of angular momentum

I w = I' w' => w' = I w/I' = (5.91)*(1.27/1.89) = 3.97 rev/s

b)

The rotational kinetic energy in each case is ½ Iw2 so the ratio of the new kinetic energy to the old is

KE'/KE = I' w'^2/(I w^2) = (1.89)(3.97)^2/(5.91)(1.27)^2 = 3.125 (no units)

Hope this helps :)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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