{"id":2167,"date":"2018-04-26T06:26:21","date_gmt":"2018-04-25T22:26:21","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=2167"},"modified":"2023-08-09T01:29:48","modified_gmt":"2023-08-09T01:29:48","slug":"moment-of-inertia-for-particle-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/moment-of-inertia-for-particle-problems-and-solutions.htm","title":{"rendered":"Moment of inertia for particle \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Moment of inertia for particle \u2013 problems and solutions<\/span><\/p>\n

1. Two balls connected by a rod, as shown in the figure below. Ignore rod’s mass<\/a>. Mass of ball P is 600 gram and mass of ball Q is 400 gram. What is the moment of inertia<\/a> of the system about AB?<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

The axis of rotation is AB.\"Moment<\/span><\/p>\n

mp<\/sub> = 600 gram = 0.6 kg, mq<\/sub> = 400 gram = 0.4 kg<\/span><\/p>\n

rp<\/sub> = 20 cm = 0.2 m, rq<\/sub> = 50 cm = 0.5 m<\/span><\/p>\n

Wanted :<\/u> The moment of inertia of the system<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

I = mp<\/sub> rp<\/sub>2<\/sup> + mq<\/sub> rq<\/sub>2<\/sup> <\/span><\/p>\n

I = (0.6 kg)(0.2 m)2<\/sup> + (0.4 kg)(0.5 m)2<\/sup><\/span><\/p>\n

I = (0.6 kg)(0.04 m2<\/sup>) + (0.4 kg)(0.25 m2<\/sup>)<\/span><\/p>\n

I = 0.024 kg m2<\/sup> + 0.1 kg m2<\/sup><\/span><\/p>\n

I = 0.124 kg m2<\/sup><\/span><\/p>\n

2. The Rod of AB with mass of 2-kg rotated about point A, the moment of inertia of the rod is 8 kg m2<\/sup>. If rotated about point O (AO = OB),what is the moment of inertia of the rod.<\/span><\/p>\n

Known :\"Moment<\/u><\/span><\/p>\n

Mass of rod AB (m) = 2 kg<\/span><\/p>\n

If rotated about point A so that the radius of rotation (r) = length of AB = r then the moment of inertia (I) = 8 kg m2<\/sup> <\/span><\/p>\n

Wanted:<\/u> If rotated about point O so that the radius of rotation (r) = length of AO = length of OB = 1\/2 r then what is the moment of inertia of the rod.<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

I = m r2<\/sup> <\/span><\/p>\n

8 kg m2<\/sup> = (2 kg) r2<\/sup><\/span><\/p>\n

8 m2<\/sup> = (2) r2<\/sup><\/span><\/p>\n

r2 <\/sup>= 8 m2<\/sup> \/ 2<\/span><\/p>\n

r2 <\/sup>= 4 m2<\/sup> <\/span><\/p>\n

r = 2 meters<\/span><\/p>\n

If rotated about point O so \u00bd r = 1 meter, then the moment of inertia :<\/span><\/p>\n

I = m r2 <\/sup>= (2 kg)(1 m)2<\/sup> = (2 kg)(1 m2<\/sup>) = 2 kg m2<\/sup> <\/span><\/p>\n

3. Two balls connected by a rod as shown in figure below. Ignore rod’s mass. What is the moment of inertia of the system.<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Mass of ball A (mA<\/sub>) = 200 gram = 0.2 kg\"Moment<\/span><\/p>\n

Mass of ball B (mB<\/sub>) = 400 gram = 0.4 kg<\/span><\/p>\n

Distance between ball A and the axis of rotation (rA<\/sub>) = 0<\/span><\/p>\n

Distance between ball B and the axis of rotation (rB<\/sub>) = 25 cm = 0.25 m <\/span><\/p>\n

Wanted :<\/u> Moment of inertia of the system<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

The moment of inertia of ball A :<\/span><\/p>\n

IA<\/sub> = (mA<\/sub>)(rA<\/sub>2<\/sup>) = (0.2)(0)2<\/sup> = 0 <\/span><\/p>\n

The moment of inertia of ball B :<\/span><\/p>\n

IB<\/sub> = (mB<\/sub>)(rB<\/sub>2<\/sup>) = (0.4)(0.25)2<\/sup> = (0.4)(0.0625) = 0.025 kg m2<\/sup> <\/span><\/p>\n

The moment of inertia of system :<\/span><\/p>\n

I = IA <\/sub>+ IB<\/sub> = 0 + 0.025 = 0.025 kg m2<\/sup> = 25 x 10-3<\/sup> kg m2<\/sup> <\/span><\/p>\n

4. Four particles with different mass, shown in figure below. Determine the moment of inertia of the system about the horizontal line P.<\/span><\/p>\n

Solution<\/span><\/p>\n

The axis of rotation is the horizontal line P.<\/i><\/span><\/p>\n

Known :\"Moment<\/u><\/span><\/p>\n

Mass of particle A (mA<\/sub>) = m <\/span><\/p>\n

Mass of particle B (mB<\/sub>) = 2m<\/span><\/p>\n

Mass of particle C (mC<\/sub>) = 3m <\/span><\/p>\n

Pass of particle D (mD<\/sub>) = 4m <\/span><\/p>\n

Distance between particle A and the axis of rotation (rA<\/sub>) = b<\/span><\/p>\n

Distance between particle B and the axis of rotation (rB<\/sub>) = b<\/span><\/p>\n

Distance between particle C and the axis of rotation (rC<\/sub>) = 2b<\/span><\/p>\n

Distance between particle D and the axis of rotation (rD<\/sub>) = 2b<\/span><\/p>\n

Wanted :<\/u> The moment of inertia of the system about the horizontal line P<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

I = mA<\/sub> rA<\/sub>2<\/sup> + mB<\/sub> rB<\/sub>2 <\/sup>+ mC<\/sub> rC<\/sub>2 <\/sup>+ mD <\/sub>rD<\/sub>2 <\/sup><\/span><\/p>\n

I = (m)(b)2 <\/sup>+ (2m)(b)2<\/sup> + (3m)(2b)2 <\/sup>+ (4m)(2b)2 <\/sup><\/span><\/p>\n

I = mb2 <\/sup>+ 2 mb2<\/sup> + (3m)(4b2<\/sup>) + (4m)(4b2<\/sup>)<\/span><\/p>\n

I = mb2 <\/sup>+ 2 mb2<\/sup> + 12 mb2 <\/sup>+ 16 mb2 <\/sup><\/span><\/p>\n

I = 31 mb2 <\/sup><\/span><\/p>\n

5. Four particles connected by a rod. Ignore rod’s mass. Determine the moment of inertia about the axis of rotation through particle m1<\/sub> and m2<\/sub>, as shown in figure below.<\/span><\/p>\n

Known<\/u><\/span><\/p>\n

Mass of particle 1 (m1<\/sub>) = 1\/4 kg \"Moment<\/span><\/p>\n

Mass of particle 2 (m2<\/sub>) = 1\/2 kg <\/span><\/p>\n

Mass of particle 3 (m3<\/sub>) = 1\/4 kg <\/span><\/p>\n

Mass of particle 4 (m4<\/sub>) = 1\/4 kg <\/span><\/p>\n

Distance between particle 1 and the axis of rotation (r1<\/sub>) = 0<\/span><\/p>\n

Distance between particle 2 and the axis of rotation (r2<\/sub>) = 0<\/span><\/p>\n

Distance between particle 3 and the axis of rotation (r3<\/sub>) = 10 cm = 10\/100 m = 1\/10 m<\/span><\/p>\n

Distance between particle 4 and the axis of rotation (r4<\/sub>) = 10 cm = 10\/100 m = 1\/10 m<\/span><\/p>\n

Wanted :<\/u> The moment of inertia<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

I = m1<\/sub> r1<\/sub>2<\/sup> + m2<\/sub> r2<\/sub>2 <\/sup>+ m3<\/sub> r3<\/sub>2 <\/sup>+ m4 <\/sub>r4<\/sub>2 <\/sup><\/span><\/p>\n

I = (1\/4)(0)2 <\/sup>+ (1\/2)(0)2<\/sup> + (1\/4)(1\/10)2 <\/sup>+ (1\/4)(1\/10)2 <\/sup><\/span><\/p>\n

I = 0 + 0 + (1\/4)(1\/100) + (1\/4)(1\/100)<\/span><\/p>\n

I = 1\/400 + 1\/400<\/span><\/p>\n

I = 2\/400<\/span><\/p>\n

I = 1\/200 kg.m2<\/sup><\/span><\/p>\n

    \n
  1. What is the moment of inertia for a particle?<\/strong><\/span>\n