From a uniform disc of radius R, a circular hole of radius \( \dfrac{R}{2}\) is cut out. The centre of the hole is at \( \dfrac{R}{2}\) from the centre of the original disc. Locate the centre of gravity of the resulting flat body

Asked by Abhisek | 1 year ago |  78

1 Answer

Solution :-

Let the mass/unit area of the original disc = σ

Radius of the original disc =2r

Mass of the original disc, m = π(2r2)σ = 4 πr2σ . . . . ... (i)

The disc with the cut portion is shown in the following figure:

NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion Q16

Radius of the smaller disc = r

Mass of the smaller disc, m’ = π r2σ

= m’ =\( \dfrac{m}{4}\)        [ From equation ( i ) ]

Let O’ and O be the respective centers of the disc cut off from the original and the original disc. According to the definition of center of mass, the center of mass of the original disc is concentrated at O, while that of the smaller disc is supposed at O′.

We know that :

OO′= \( \dfrac{R}{2}\) = r2

After the smaller circle has been cut out, we are left with two systems whose masses are:

m’ (\( \dfrac{m}{4}\)) concentrated at O′, and m (concentrated at O).

(The negative sign means that this portion has been removed from the original disc.)

Let X be the distance of the center of mass from O.

We know :

X = \( \dfrac{ (m_1 r_1 + m_2 r_2) }{ (m_1 + m_2)}\)

X =\( \dfrac{m × 0 – m’ \times \dfrac{r}{2}}{( M + (-M‘)}\)

\( \dfrac{-R}{6}\)

(The negative sign indicates that the center of mass is\( \dfrac{R}{6}\) towards the left of O.)

Answered by Pragya Singh | 1 year ago

Related Questions

Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass:

(i) Show  p = p’i + miV

Where pi is the momentum of the ith particle (of mass mi ) and p’i = mivi‘.  Note v’i is the velocity of the ith particle with respect to the centre of mass.

Also, verify using the definition of the centre of mass that Σp’i = 0

(ii) Prove that K = K′ + \( \dfrac{1}{2}\)MV2

Where K is the total kinetic energy of the system of particles, K′ is the total kinetic energy of the system when the particle velocities are taken relative to the centre of mass and MV2 /2 is the kinetic energy of the translation of the system as a whole.

(iii) Show L = L’+ R × MV 

where L’ = ∑r’i × p’i is the angular momentum of the system about the centre of mass with velocities considered with respect to the centre of mass. Note r’i = ri – R, rest of the notation is the standard notation used in the lesson. Note L’ and  MR × V can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.

(iv) Prove that : 

\( \dfrac{dL’}{dt}\)= ∑ r’i x \( \dfrac{dp’}{dt}\)

Further prove that :

\( \dfrac{dL’}{dt}\) = τ’ext

Where τ’ext is the sum of all external torques acting on the system about the centre of mass. ( Clue : Apply  Newton’s Third Law and  the definition of centre of mass . Consider that internal forces between any two particles act along the line connecting the particles.)

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