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Figure shows a rigid structure consisting of a circular hoop of radius Rand massm, and a square made of four thin bars, each of lengthRand massm. The rigid structure rotates at a constant speed about a vertical axis, with a period of rotation of2.5s. AssumingR=0.5mandrole="math" localid="1660971946053" m=2.0kg,

(a) Calculate the structure’s rotational inertia about the axis of rotation?

(b) Calculate its angular momentum about that axis?

Short Answer

Expert verified
  1. The total rotational inertia of the rigid structure about the axis of rotation is1.6kg.m2.
  2. The angular momentum the rigid structure is4.0kg.m2s.

Step by step solution

01

Given

  1. The mass of the hoop,m=2kg .
  2. The radius of the hoop,R=0.5m.
  3. The mass of the bar, m=2kg.
  4. The length of the bar,R=0.5m.
  5. The period of the rotation of the rigid structure, T=2.5s.
02

To understand the concept

Find the rotational inertia of the rigid structure by adding the rotational inertia of the circular hoop and the square about the axis of rotation. Then by using the given period of rotation of the structure, we can find its angular speed. By using the relation between ³¢,±õ²¹²Ô»åÓ¬we can find the angular momentum of the rigid structure.

Formula:

i. The M.I of the circular hoop along the axis passing through its center is,I=12mR2

ii. The M.I of the bar (rod) along the axis passing through its center is,I=ml212

iii. Parallel axis theorem is,I=Icom+mh2

iv. The angular momentum of system isL=±õÓ¬

03

Step 3: Calculate the structure’s rotational inertia about the axis of rotation

(a)

The rotational inertia of the circular hoop is

I1=Icom+mh2⇒I1=12mR2+mR2⇒I1=32mR2

The square is made up of four bars, out of which one is along the axis of rotation. So its moment of inertia is zero.

I2=0

Therotational inertiaof the other side which is parallel to the axis of rotation is

I=Icom+mh2⇒I3=0+mR2⇒I3=mR2

Therotational inertiaof the bars which are perpendicular to the axis of rotation are same and are

I4=I5=112mR2+mR22⇒I4=I5=13mR2

Therefore, the total rotational inertia of the rigid structure is

I=I1+I2+I3+I4+I5⇒I=32mR2+0+mR2+13mR2+13mR2⇒I=196=1.66~1.6kgm2

04

 Calculate its angular momentum about that axis

(b)

The angular speed of the rigid structure is

Ó¬=»åθdt⇒Ӭ=2Ï€2.5⇒Ӭ=2.5rads

The angular momentum of the rigid structure is,

L=±õӬ⇒L=1.6×2.5⇒L=4.0kgm2s

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