Question 71:
moderateThree masses are placed on the $x$-axis : $300\text{ g}$ at origin, $500\text{ g}$ at $x = 40\text{ cm}$ and $400\text{ g}$ at $x = 70\text{ cm}$. The distance of the center of mass from the origin is :
(2012 Mains)
Use the centre of mass formula $x_{cm} = \frac{\sum m_i x_i}{\sum m_i}$. Substituting the given values: $x_{cm} = \frac{300(0) + 500(40) + 400(70)}{300 + 500 + 400} = \frac{48000}{1200} = 40\text{ cm}$. Option (a) is correct.
Question 72:
moderateTwo persons of masses $55\text{ kg}$ and $65\text{ kg}$ respectively, are at the opposite ends of a boat. The length of the boat is $3.0\text{ m}$ and weighs $100\text{ kg}$. The $55\text{ kg}$ man walks up to the $65\text{ kg}$ man and sits with him. If the boat is in still water the center of mass of the system shifts by:
(2012 Pre)
Since no external horizontal force acts on the system (boat + persons), the position of the centre of mass of the system remains unchanged. Thus, the shift in the centre of mass is zero. Option (c) is correct.
Question 73:
moderateA man of $50\text{ kg}$ mass is standing in a gravity free space at a height of $10\text{ m}$ above the floor. He throws a stone of $0.5\text{ kg}$ mass downwards with a speed $2\text{ m/s}$. When the stone reaches the floor, the distance of the man above the floor will be :
(2010 Pre)
In gravity-free space, no external force acts, so the centre of mass remains at its initial height of $10\text{ m}$. Using COM conservation: $M_m h_m + M_s h_s = (M_m + M_s) Y_{cm} \implies 50(h) + 0.5(0) = (50 + 0.5)(10) \implies h = 10.1\text{ m}$. Option (b) is correct.
Question 74:
easyA uniform circular disc of radius $50\text{ cm}$ at rest is free to turn about an axis which is perpendicular to its plane and passes through its center. It is subjected to a torque which produces a constant angular acceleration of $2.0\text{ rad s}^{-2}$. Its net acceleration in $\text{ms}^{-2}$ at the end of $2.0\text{ s}$ is approximately:
(2016-I)
Concept: Combination of tangential and centripetal accelerations. Formula: $a = \sqrt{a_c^2 + a_t^2}$. Solution: $a_t = r\alpha = 1.0$, $a_c = \omega^2 r = 8.0$, giving $a = \sqrt{8^2 + 1^2} \approx 8.0\text{ ms}^{-2}$.
Question 75:
easyTwo particles which are initially at rest, move towards each other under the action of their internal attraction. If their speeds are $v$ and $2v$ at any instant, then the speed of centre of mass of the system will be:
(2010 Pre)
Concept: Since external force on the system is zero, the acceleration of the center of mass is zero. Formula: $v_{cm} = \frac{\sum m_i v_i}{\sum m_i}$. Solution: Since the system starts from rest and only internal forces act, velocity of center of mass remains zero.
Question 76:
easyTwo bodies of mass $1text{ kg}$ and $3text{ kg}$ have position vectors $\hat{i} + 2\hat{j} + \hat{k}$ and $-3\hat{i} – 2\hat{j} + \hat{k}$, respectively. The center of mass of this system has a position vector:
(2009)
Concept: Center of mass position vector formula. Formula: $\vec{r}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2}$. Solution: Substituting the given masses and position vectors yields $-2\hat{i} - \hat{j} + \hat{k}$.
Question 77:
easyConsider a system of two particles having masses $m_1$ and $m_2$. If the particle of mass $m_1$ is pushed towards the mass centre of particles through a distance ‘$d$’ by what distance would the particle of mass $m_2$ move so as to keep the mass centre of particles at the original position:
(2004)
Concept: Shift in center of mass must be zero. Formula: $m_1 \Delta x_1 = m_2 \Delta x_2$. Solution: Substituting $\Delta x_1 = d$ gives $\Delta x_2 = \frac{m_1}{m_2}d$.
Question 78:
easyThe centre of mass of system of particles does not depend on:
(1997)
Concept: Definition and properties of center of mass. Formula: $\vec{R}_{cm} = \frac{\sum m_i \vec{r}_i}{\sum m_i}$.
Solution: Center of mass depends only on masses and positions, independent of internal or external forces acting on particles.