Question

find the distance of centre of mass of the uniform rod of length πr/3

Answer 1

we know, centre of mass of rod is the middle point of rod.

if length of rod is L , then centre of mass of rod exists at L/2.

here length of rod is πr/3

so, centre of mass of rod is at πr/6

Answer 2

Answer:

Centre of Mass, Cm ⇒ $\frac{\pi \, r}{6}$

Explanation:

For the given condition two cases arise;-

[CASE 1] (If the centre of mass asked is for a discrete system of particles)

Let R₁(0, 0) & R₂(L, 0) be the origin and other position in the x-axis. Let m be the mass for both the positioned particles ( since rod is uniform ).

Then, we know that by formula;-

∵       Xcm = x₁r₁ + x₂r₂ / m₁ + m₂

Note; Xcm represents centre of mass in x-axis.

∴       Xcm = R₁(0) + R₂(m)/ m + m

          Xcm = R₂m / 2m

          Xcm = R₂/ 2

          Xcm = L/2    [Given]

Now, according to the question, L = πr/3, so;-

          Xcm = πr/3 / 2

          Xcm = πr/3 × 2

         Xcm = πr/6

[CASE 2] (If the centre of mass asked is for a continuous system of particles)

Refer to the attached images for case 2's diagram.

We find the centre of mass for small mass, distance after travelling x distance in x-axis with the help of integration.

We know that, by formula;

            $Xcm \, = \frac{\int\limits^L_0 {xdm} }{\int\limits^L_0 {dm} }$

Since, $\frac{dm}{dx}\, = Lambda$, Hence, $dm = dx \, Lambda$

            $Xcm = \frac{\int\limits^L_0 {xdx \,Lambda} }{\int\limits^L_0 {dx \, Lambda} }$

Since, Lambda (λ) = constant, hence;-

           $Xcm = \frac{Lambda \, \int\limits^L_0 {xdx} }{Lambda \, \int\limits^L_0 {dx} }$

           $Xcm = \frac{\int\limits^L_0 {dx^{2} } }{\int\limits^L_0 {dx} }$

           $Xcm = [\frac{x^{2} }{2x} ]^{2}$

           $Xcm = \frac{L}{2}$

Now, according to the question, L = πr/3, so;-

           $Xcm = \frac{\frac{\pi \, r}{3} }{2}$

           $Xcm = \frac{\pi r}{6}$

Hence, in both the cases, the value of centre of mass is same i.e. Xcm = $\frac{\pi r}{6}$

Hope it helps! ;-))