Two masses are joined together by a massless rod of length 0.8m. One mass is 1kg

Question

Two masses are joined together by a massless rod of length 0.8m. One mass is 1kg and the other is 3kg. They are rotating with a period of 0.25s.

A) What is their moment of inertia and what is their angular momentum?

B) A hole is drilled in the massless rod system. This hole is 0.2m from the larger mass. If a peg is placed through the hole and the system is now rotating around this peg at the same period, what is its new moment of inertia and angular momentum?

C) Repeat part A, but the mass of the rod now has a uniform linear mass density of 400 grams per meter.

D) Repeat part B, but the mas of the rod now has a uniform linear mass density of 200 grams per meter.

Explanation / Answer

The length of rod, L = 0.8 m masses m1 = 1 kg ; m2 = 3 kg ; The time period, T = 0.25 s Let the centre of mass of the system is at a distance ' x ' from m1 and 0.8-x from m2 Then, m1x = m2(0.8 - x) m1x + m2x = m2 * 0.8 From the above, x = 0.6 m So the system is rotating in circular part with the radius 0.6m So the moment of inertia, I = mr^2 = (m1 + m2)r^2 = 1.44 kg m^2 The angular velocity, ? = 2p/T = 25.13 rad/s The angular momentum, L = I? = 36.18 Js b. Conservation of angular momentum [ 1 * ( 0.8 /2 ) ^2 + 3 * ( 0.8 / 2 ) ^2 ] ( 2 p / 0.25 ) = [ 1 * ( 0.8 /2 ) ^2 + 3 * ( 0.8 / 2 ) ^2 ] ( 2 p / 0.25 ) new I = [ 3 * 0.2^2 + 1 * ( 0.8 - 0.2 )^2 ] = 0.48 ; L = I ? = 0.48 * ( 2p / 0.25 ) = 12.0637 ----------------------------------------------------------------------------------------- The linear density = 400gm/m So the mass of given rod, m3 = 400gm * 0.8 = 320 gm = 0.32 kg Then the moment of inertia, I = (m1 + m2 + m3)r^2 = 1.56 kg m^2 The angular momentum, L = I? = 39.2 Js

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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