A large horizontal circular platform (M = 109.2 kg, r = 3.19 m) rotates about a

Question

A large horizontal circular platform (M = 109.2 kg, r = 3.19 m) rotates about a frictionless vertical axle. A student (m = 57.42 kg) walks slowly from the rim of the platform toward the center. The angular velocity go of the system is 2.22 rad/s when the student is at the rim. Find the moment of inertia of platform through the center with respect to the z-axis. Tries 0/5 Find the moment of inertia of the student about the center axis (while standing at the rim) of the platform. Tries 0/5 Find the moment of inertia of the student about the center axis while the student is standing 1.89 m from the center of the platform. Tries 0/5 Find the angular speed when the student is 1.89 m from the center of the platform. Tries 1/5

Explanation / Answer

(1)

Moment of inertia of the platform about z-axis:

I1 = M r^2 / 2

   = (109.2 * 3.19^2 ) / 2

    = 555.62 kg.m^2

    =5.56 x 10^2 kg.m^2

(2)

Moment of inertia of the student while standing at the rim:

I2 = m r^2

   = 57.42 * 3.19^2

   = 584.3 kg.m^2

   = 5.84 x 10^2 kg.m^2

(3)

Moment of inertia of the student while standing at the 1.89 from the centre of platform

I3 = m r^2

   = 57.42 * 1.89^2

   = 205.1 kg.m^2

   = 2.05 x 10^2 kg.m^2

(4)

Angular speed of the student while standing at the 1.89 from the centre of platform.

From law of conservation of angular momentum,

( I1 + I2 ) w1 = ( I1 + I3 ) w2

( 555.62 + 584.3 ) * 2.22 = ( 555.62 + 205.1 ) * w2

2530.6 = 760.72 w2

w2 = 3.33 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.