JEE Advanced PYQ System of Particles and Rotational Motion

Q 1 :

Consider regular polygons with number of sides $n = 3, 4, 5, \dots$ as shown in the figure. The center of mass of all the polygons is at height $h$ from the ground. They roll on a horizontal surface about the leading vertex without slipping and sliding as depicted. The maximum increase in height of the locus of the center of mass for each polygon is $Δ$ . Then $Δ$ depends on $n$ and $h$ as [2017]

[IMAGE 152]

$Δ = h \sin^{2} (\frac{π}{n})$
$Δ = h (\frac{1}{\cos (\frac{π}{n})} - 1)$
$Δ = h \sin (\frac{2 π}{n})$
$Δ = h \tan^{2} (\frac{π}{2 n})$

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(2)

[IMAGE 153]

When the polygon rolls without slipping about the point $' P'$ . The point O reaches the maximum height (Point O' in the figure) by moving in a circle of radius $' r'$ about the point 'P'.

So, maximum increase in height of centre of mass 'O' is given by $Δ = r - h = \frac{h}{\cos (\frac{π}{n})} - h$

$\Rightarrow Δ = h (\frac{1}{\cos (\frac{π}{n})} - 1)$

Q 2 :

Look at the drawing given in the figure which has been drawn with ink of uniform line-thickness. The mass of ink used to draw each of the two inner circles, and each of the two line segments is $m$ . The mass of the ink used to draw the outer circle is $6 m$ .

The coordinates of the centres of the different parts are: outer circle (0, 0), left inner circle $(- a, a),$ right inner circle $(a, a)$ , vertical line (0, 0) and horizontal line $(0, - a)$ . The $y$ -coordinate of the centre of mass of the ink in this drawing is [2009]

[IMAGE 154]

$\frac{a}{10}$
$\frac{a}{8}$
$\frac{a}{12}$
$\frac{a}{3}$

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(1)

The drawing given in the figure is made up of five bodies i.e., three circles and two straight lines of uniform mass distribution or we can assume the system to be made up of five point masses where the mass of each is considered at its geometrical centre.

[IMAGE 155]

The $y$ -coordinate of the centre of mass is

$y_{c m} = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3} + m_{4} y_{4} + m_{5} y_{5}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5}}$

$∴$ $y_{c m} = \frac{6 m \times 0 + m \times 0 + m \times a + m \times a + m (- a)}{6 m + m + m + m + m}$

$= \frac{m a}{10 m} = \frac{a}{10}$

Q 3 :

A block of mass M has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surface of a fixed table. Initially the right edge of the block is at $x = 0$ , in a co-ordinate system fixed to the table. A point mass $m$ is released from rest at the topmost point of the path as shown and it slides down.

When the mass loses contact with the block, its position is $x$ and the velocity is $v$ . At that instant, which of the following options is/are correct? [2017]

[IMAGE 156]

The position of the point mass $m$ is: $x = - \sqrt{2} \frac{m R}{M + m}$
The velocity of the point mass $m$ is: $v = \sqrt{\frac{2 g R}{1 + \frac{m}{M}}}$
The $x$ -component of displacement of the center of mass of the block M is: $- \frac{m R}{M + m}$
The velocity of the block M is: $V = - \frac{m}{M} \sqrt{2 g R}$

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Select one or more options

(2, 3)

Let the block of mass M moves by distance $x$ towards left.

$M x = m (R - x)$

$\Rightarrow x = \frac{m R}{M + m} towards left$ $∴$ $x = - \frac{m R}{M + m}$

If $v$ is the velocity of mass $' m'$ as it leaves the block and $V$ is the velocity of block at that instant then according to conservation of linear momentum

$m v = M V$

By energy conservation

$m g R = \frac{1}{2} m v^{2} + \frac{1}{2} M V^{2}$

Solving we get, $v = \sqrt{\frac{2 g R}{1 + \frac{m}{M}}} and V = \frac{m}{M} \sqrt{\frac{2 g R}{1 + \frac{m}{M}}}$

Q 4 :

STATEMENT-1: If there is no external torque on a body about its center of mass, then the velocity of the center of mass remains constant.

STATEMENT-2: The linear momentum of an isolated system remains constant. [2007]

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
Statement-1 is True, Statement-2 is False
Statement-1 is False, Statement-2 is True

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(4)

There is no external torque on a body about its centre of mass, so no Ven is constant. For velocity of centre of mass to remain constant the net force acting on a body must be zero.

The linear momentum of an isolated system remains constant.

Topic Question Set

Q 4 :

STATEMENT-1: If there is no external torque on a body about its center of mass, then the velocity of the center of mass remains constant.

STATEMENT-2: The linear momentum of an isolated system remains constant. [2007]

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Topic Question Set

Q 4 : STATEMENT-1: If there is no external torque on a body about its center of mass, then the velocity of the center of mass remains constant. STATEMENT-2: The linear momentum of an isolated system remains constant. [2007]

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Q 4 :

STATEMENT-1: If there is no external torque on a body about its center of mass, then the velocity of the center of mass remains constant.

STATEMENT-2: The linear momentum of an isolated system remains constant. [2007]