Centre of mass

Q: A block of mass M with a semicircular track of radius R rest on a horizontal frictionless surface. A uniform cylinder of radius r and mass m is released from rest at the top point A. The cylinder slips in the semicircular frictionless track. How far the block moved when the cylinder reaches the bottom of the track?

Sol: Since no net horizontal force acts on the system, it conserves its horizontal momentum.

$\displaystyle \vec{P} = M \vec{v_b} + m \vec{v_c} = 0 $ .

Because initially the system was stationary.

$\displaystyle M \vec{v_b} + m \vec{v_c} = 0 $ , Where $\vec{v_b}$ & $\vec{v_c}$ are the horizontal component of velocities of the block and the cylinder respectively

v_cb = velocity of the cylinder relative to the block = v (say)

$\displaystyle v_b = \frac{m v_{cb} sin\theta}{M + m}$ ( Numerically )

$\displaystyle \int_{0}^{t} v_b dt = \frac{m}{M+m}\int_{0}^{t} (v_{cb})_x dt $

where t = time taken by the cylinder to arrive at the bottom of the block& (v_cb)x = horizontal component of velocity of the cylinder relative to the block.

$\displaystyle x_b = \frac{m x_{cb}}{M+m} $ ; putting x_bc = (R-r) we obtain

$\displaystyle x_b = \frac{m}{M+m}(R-r) $ ; Where $\displaystyle \int_{0}^{t} (v_{cb})_x dt = (R-r)$ and $\displaystyle \int_{0}^{t} v_b dt = x_b $

Q: Two particles A and B of which lighter particle has mass m, are released from infinity. They move towards each other under their mutual force of attraction. If their speeds are v and 2 v respectively find the K.E. of the system.

Sol: Since the particles are moving under their mutual gravitation, net force acting on the system (m₁ + m₂) is equal to $\displaystyle \vec{F_1} + \vec{F_2} = 0$

Therefore the linear momentum of the system is constant

$\displaystyle | m_1\vec{v_1} + m_2 \vec{v_2} | = constant$

Since initially the particles were released from rest, P_i = 0

$\displaystyle P_f = m_1\vec{v_1} + m_2 \vec{v_2} = 0 $

$\displaystyle m_2 = \frac{m_1 v_1}{v_2}$ (numerically)

$\displaystyle m_2 = \frac{m v}{2 v} = \frac{m}{2}$ ; where m₁ = m , v₁ = v & v₂ = 2v

The KE of the system $K.E = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$

$\displaystyle K.E = \frac{1}{2}m v^2 + \frac{1}{2}(m/2 ) (2v)^2$

$\displaystyle K.E = \frac{3}{2}m v^2 $

q: The center of mass of a solid hemisphere of radius 8 cm is x cm from the center of the flat surface . Then value of x is ….

The coordinate of center of mass of a uniform flag shaped lamina (thin flat plate ) of mass 4 kg (the coordinate of same are shown in figure ) are

(a) (1.25 m , 1.50 m)

(b) (0.75 m , 0.75 m)

(c) (0.75 m , 1.75 m )

(d) (1 m , 1.75 m)