Perpendicular Axis Theorem
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Analogue of mass in rotational motion is:
Moment of inertia
Angular momentum
Gyration
None of these
Let 'I' be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle θ with AB. The moment of inertia of the plate about the axis CD is then equal to
- I
- Isin2θ
- Icos2θ
- Icos2(θ2)
- ρL38π2
- ρL316π2
- 5ρL316π2
- 3ρL38π2
- MR2
- 2MR2
- MR24
- MR22
- I1+I2=I3+I4
- I1+I4=I2+I3
- I1+I3=I2+I4
- I1−I3=I2−I4
- 20 kg m2
- 5 kg m2
- 10 kg m2
- 40 kg m2
The moment of inertia of a thin square plate ABCD of uniform thickness about an axis passing through the centre O and perpendicular to the plane of the plate can NOT be given by :
where, I1, I2, I3, I4 are the moments of inertia of the plates about axes 1, 2, 3, 4 respectively.
- I1+I2
- I3+I4
- I1+I3
- I1+I2+I3+I4
- 20 kg m2
- 5 kg m2
- 10 kg m2
- 40 kg m2
- 9MR2
- 3MR2
- √27MR2
- √45MR2
- 3MR2
- 32MR2
- 5MR2
- 72MR2
- I
- Isin2θ
- Icos2θ
- Icos2θ2
Statement 1:Iz=Ix+Iy
Statement 2:Ix=Iz+Iy and Iy=Ix+Iz
where x, y and z are the axes of rotation of the body.
Choose the correct option.
- Statement 2 is correct and statement 1 is incorrect.
- Statement 1 is correct and statement 2 is incorrect.
- Both statements are correct
- None of these
- 6 kg-m2
- 4 kg-m2
- 8 kg-m2
- 10 kg-m2
- 3mr2
- 165mr2
- 4mr2
- 115mr2
- 23Ml2
- 133Ml2
- 13Ml2
- 43Ml2
Two uniform identical rods each of mass M and length l are joined to form a cross as shown in figure. Find the moment of inertia of the cross about a bisector as shown dotted in the figure.
ML26
ML224
ML212
ML23
- I
- Isin2θ
- Icos2θ
- Icos2θ2
- 2 kg m2
- 4 kg m2
- 6 kg m2
- 8 kg m2
- 23Ml2
- 133Ml2
- 13Ml2
- 43Ml2
- 75 kg m2
- 45 kg m2
- 15 kg m2
- 55 kg m2
- √2IAC=IEF
- IAC=2IEF
- IAC=IEF
- IAC=√2IEF
- IAB=1256 kg.m2
- IAB=125√32 kg.m2
- IAB=1253 kg.m2
- IAB=2503 kg.m2
- ρL38π2
- ρL316π2
- 5ρL316π2
- 3ρL38π2