Calculate the radius of gyration of a hoop of
radius 5.00 cm rotating about an axis passing through its centre and normal to its plane.
Answers
Answer:
radius of gyration of a hoop √35 cm
Explanation:
The radius of gyration of a hoop of radius 5.00 cm rotating about an axis passing through its centre and normal to its plane.
I(tangent) = 2/5 Ma² + MR²
∴ I(tangent) = 7/5 MR²
∴ Mk² = 7/5 MR²
∴ k = √(7/5 x 5² = √35 cm
Answer:
FOR ALL NIOS NOTES || ASSIGNMENT || PRACTICAL FILE || ADMISSION CALL US : 9953056766
Facebook : https://www.facebook.com/MissionGuideInstitute/
YouTube :https://www.youtube.com/channel/UCCxRi9ksnVInxJPz3f3JYSQ
Explanation:
Radius of gyration is defined as the distance from the axis of rotation to a point where the total mass of the body is supposed to be concentrated, so that the moment of inertia about the axis may remain the same. Simply, gyration is the distribution of the components of an object.
The radius of gyration of a hoop of radius 5.00 cm rotating about an axis passing through its centre and normal to its plane.
I(tangent) = 2/5 Ma² + MR²
∴ I(tangent) = 7/5 MR²
∴ Mk² = 7/5 MR²
∴ k = √(7/5 x 5² = √35 cm
Ques 4 : Find the general solution for x in
Cos 3x + Cos x - Cos 2x = 0 (i.e. zero)