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>> $Rigid Body and Motion of Rigid Body - II$

Rigid Body and Motion of Rigid Body - II

Q: The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16sec.? find.(i) Angular acceleration(ii) Assuming acceleration to be uniform how many rotations does the engine make this bike?(iii) State law of conservation of linear momentum?

\${\omega _1} = 1200\,rpm\$ \$ = {{1200 \times 2\pi } \over {60}} = 40\pi \,rad.{s^{ - 1}}\$ \${\omega _2} = {{3120 \times 2\pi } \over {60}} = 10\pi \,rad.{s^{ - 1}}\$ \$t = 16\,\sec \$ \$(1)\$ Angular acceleration \$a = {{({\omega _2} - {\omega _1})} \over {16}} = {{64\pi } \over {16}}\$ \$ = 4\pi \,rad.{s^{ - 1}}\$ \$(2)\$ Angle, \$\theta\$ \$ = 40\pi \,16 + {1 \over 2}(4\pi ) \times 256\$ \$ = 1152\,rad.\$ \$(3)\$ When there is no any external force applied on a system of partials then, momentum of system of particles will be consecrate or you can say that if external force is zero, then momentum of system of particles is initially \$ = \$ Momentum of system of particles is finally.

Q: A T-shaped object with dimensions shown in the figure is lying on a smooth floor. A force \$\vec { F }\$ is applied at the point P parallel to AB, such that the object has only the translational motion without rotation. Find the location of P with respect to C.

Given: A T-shaped object with dimensions shown in the figure is lying on a smooth floor. A force \$\vec F\$ is applied at the point P parallel to AB, such that the object has only the translational motion without rotation. To find the location of P with respect to C.Solution:To have the required motion, the force \$\vec F\$ has to be applied at the center of mass.The point P must be the centre of mass of the T-shaped object since the force F does not produce any rotational motion of the object.The horizontal part of the T-shaped object has length L. If the mass of the horizontal portion is ‘m’, the mass of the vertical portion of the T- shaped object is 2m since its length is 2L. Therefore, the T-shaped object reduces to two point masses m and 2m at distances 2L and L respectively from the point C. The distance ‘r’ of the centre of mass of the system from the point C is given by\$r=\dfrac {m_1r_1+m_2r_2}{m_1+m_2}\\\implies r=\dfrac {m\times 2L+2m\times L}{m+2m}\\\implies r=\dfrac {4L}{3}\$is the location of P with respect to C.

Q: State the condition of equilibrium of a system of coplanar parallel forces.

The condition of equilibrium of a body acted upon by a number of parallel forces is, 1. Algebraic sum of the forces must be zero. i.e., Sum of the forces acting in the upward direction must be equal to the sum of the forces acting in the downward direction.2.The algebraic sum of the moments of the forces about any point in their plane must be zero. i.e., Sum of the moments of the forces about a point in the anticlockwise direction is equal to the sum of the moments of the forces about the same point in the clockwise direction.

Q: A uniform circular disc of radius is taken. A circular portion of radius b has been removed from its as shown in the figure. If the centre of hole is at a distance c from the centre of the disc, the distance \$x_2\$ of the centre of mass of the remaining part from the initial centre of mass O is given by :

\$x_{cm} = \dfrac{A_1x_1 - A_2 x_2}{A_1 - A_2}\$\$A_1 = \pi \times a^2 \$ {area of whole disc}\$A_2 = \pi b^2\$ {area of remove d disc}\$x_1 = 0\$ (centre of mass of whole disc)\$x_2 = c\$ (centre of mass of removed disc)\$x_{cm} = \dfrac{(xa^2) (0) - (\pi b^2) (c)}{(\pi a^2 - \pi b^2)} = \dfrac{- \pi b^2 c}{\pi (a^2 - b^2)}\$\$x_{cm} = \dfrac{-b^2 c}{(a^2 - b^2)}\$ From point 0, centre of mass l is at left at a distance \$= \dfrac{b^2 c}{(a^2 - b^2)}\$Hence, option \$(B)\$ is correct answer.

Q: A uniform metre scale is in equilibrium position. Calculate the mass of the ruler.

\$\tau\$ = \$r\times F\$ Let the mass of the ruler be \$M\$ grams.Centre of mass of the ruler is at it's center.Therefore, centre of mass of the ruler is \$20\$ cm away from the hinge . It is given that the ruler is in equilibrium position.Balancing the torques due to the mass of the rod and due to the 40 grammass suspended from the ruler about the hinge :\$40\$*\$25\$ = \$M\$*\$20\$\$M\$ = \$50\$ grams.Hence , the mass of the ruler is \$50\$ grams .

Q: A uniform metre rule which has a mass of \$75\ g\$ is suspended by a thread at the \$75\ cm\$ mark and is balanced by a mass of weight W hanging from the \$90\ cm\$ mark. Calculate the value of W.

from Equilibrium,clockwise moment=anticlockwise moment\$W \times 15 = 75 \times g \times 25\$\$m \times g \times 15 = 75 \times g \times 25 \$\$\therefore = \cfrac{ 75 \times 25 }{ 15 }=125 g\$

Q: A thin uniform rod \$AB\$ of mass \$m= 1 kg\$ moves translationally with acceleration \$a =2 m/s^{2}\$ due to two anti parallel forces \$ F_{1}\$ and \$F_{2}\$ The distance between the points at which these forces are applied is equal to \$ l= 20 cm.\$ Besides, it is known that \$F_{2} = 5N\$. Find the length of the rod (in meteres).

Let \$x\$ be the distance of centre \$C\$ of rod from point \$D.\$ Then\$F_{2} -F_{1} =ma\$ or \$F_{1} =3N\$Further, \$ T_{c}=0 \$\$\therefore F_{2}x=F_{1} (0.2+x) \$\$ 5x=F_{1} (0.2 +x) \$ \$ \therefore 5x=3 (0.2 +x) \$or \$ x=0.3 m\$ \$\displaystyle \therefore \$ Length of rod \$=2(x+0.2) =1.0 m\$

Q: A uniform rod of mass \$15 kg\$ is held stationary with the help of a light string as shown in fig. \$L=5m\$, angle between rod and ground is \$37^0\$. The tension in the string is :

For the system to be in equilibrium, moment about the point of hinge must be zero.\$\implies mg\dfrac{L}{2}\cos\theta-T(3m)=0\$For,\$g=10m, \theta=37^{\circ},L=5m\$\$T=100N\$

Q: A uniform metre scale balances horizontally on a knife edge placed at \$55\ cm\$ mark. When a mass of \$25\ g\$ is supported from one end, then the mass of the scale is :

The centre of gravity of a meter scale will be acting at the mid point (50 cm) of the scale. For the balance, the moment of forces on both sides of knife edge will be equal.So, \$mg\times (55-50)=25g\times (100-55)\$ or \$m=225 gram\$

Q: What torque will increase angular velocity of a solid disc of mass \$16kg\$ and diameter \$1m\$ from zero to \$2\$rpm in \$8s\$?

Torque \$\left( \tau \right) =I\alpha \$\$\tau =\cfrac { 1 }{ 2 } \times M{ R }^{ 2 }\times \cfrac { 2\pi \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ t } \$\$\therefore\$ \$\tau =16\times { \left( \cfrac { 1 }{ 2 } \right) }^{ 2 }\times \pi \cfrac { \left( 2-0 \right) }{ 8 } =\pi N-m\$

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REVISE WITH CONCEPTS

What Kind of a Motion Rigid Body Can Have?

ExampleDefinitionsFormulaes

Important Questions

The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16sec.? find.
(i) Angular acceleration
(ii) Assuming acceleration to be uniform how many rotations does the engine make this bike?
(iii) State law of conservation of linear momentum?

Medium

View solution

A T-shaped object with dimensions shown in the figure is lying on a smooth floor. A force $F$ is applied at the point P parallel to AB, such that the object has only the translational motion without rotation. Find the location of P with respect to C.

Hard

View solution

State the condition of equilibrium of a system of coplanar parallel forces.

Medium

View solution

A uniform circular disc of radius is taken. A circular portion of radius b has been removed from its as shown in the figure. If the centre of hole is at a distance c from the centre of the disc, the distance $x_{2}$ of the centre of mass of the remaining part from the initial centre of mass O is given by :

Hard

View solution

A uniform metre scale is in equilibrium position. Calculate the mass of the ruler.

Medium

View solution

A uniform metre rule which has a mass of $75 g$ is suspended by a thread at the $75 c m$ mark and is balanced by a mass of weight W hanging from the $90 c m$ mark. Calculate the value of W.

Medium

View solution

A thin uniform rod $A B$ of mass $m = 1 k g$ moves translationally with acceleration $a = 2 m / s^{2}$ due to two anti parallel forces $F_{1}$ and $F_{2}$ The distance between the points at which these forces are applied is equal to $l = 20 c m .$ Besides, it is known that $F_{2} = 5 N$ . Find the length of the rod (in meteres).

Hard

View solution

A uniform rod of mass $15 k g$ is held stationary with the help of a light string as shown in fig. $L = 5 m$ , angle between rod and ground is $3 7^{0}$ . The tension in the string is :

Medium

View solution

A uniform metre scale balances horizontally on a knife edge placed at $55 c m$ mark. When a mass of $25 g$ is supported from one end, then the mass of the scale is :

Medium

View solution

What torque will increase angular velocity of a solid disc of mass $16 k g$ and diameter $1 m$ from zero to $2$ rpm in $8 s$ ?

Medium

View solution

Rigid Body and Motion of Rigid Body - II

Language

Rate

REVISE WITH CONCEPTS

What Kind of a Motion Rigid Body Can Have?

The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16sec.? find. (i) Angular acceleration (ii) Assuming acceleration to be uniform how many rotations does the engine make this bike? (iii) State law of conservation of linear momentum?

A T-shaped object with dimensions shown in the figure is lying on a smooth floor. A force F is applied at the point P parallel to AB, such that the object has only the translational motion without rotation. Find the location of P with respect to C.

State the condition of equilibrium of a system of coplanar parallel forces.

A uniform metre scale is in equilibrium position. Calculate the mass of the ruler.

A uniform metre rule which has a mass of 75 g is suspended by a thread at the 75 cm mark and is balanced by a mass of weight W hanging from the 90 cm mark. Calculate the value of W.

A thin uniform rod AB of mass m=1kg moves translationally with acceleration a=2m/s2 due to two anti parallel forces F1​ and F2​ The distance between the points at which these forces are applied is equal to l=20cm. Besides, it is known that F2​=5N. Find the length of the rod (in meteres).

A uniform rod of mass 15kg is held stationary with the help of a light string as shown in fig. L=5m, angle between rod and ground is 370. The tension in the string is :

A uniform metre scale balances horizontally on a knife edge placed at 55 cm mark. When a mass of 25 g is supported from one end, then the mass of the scale is :

What torque will increase angular velocity of a solid disc of mass 16kg and diameter 1m from zero to 2rpm in 8s?

The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16sec.? find.
(i) Angular acceleration
(ii) Assuming acceleration to be uniform how many rotations does the engine make this bike?
(iii) State law of conservation of linear momentum?

A T-shaped object with dimensions shown in the figure is lying on a smooth floor. A force $F$ is applied at the point P parallel to AB, such that the object has only the translational motion without rotation. Find the location of P with respect to C.

A uniform metre rule which has a mass of $75 g$ is suspended by a thread at the $75 c m$ mark and is balanced by a mass of weight W hanging from the $90 c m$ mark. Calculate the value of W.

A uniform rod of mass $15 k g$ is held stationary with the help of a light string as shown in fig. $L = 5 m$ , angle between rod and ground is $3 7^{0}$ . The tension in the string is :

A uniform metre scale balances horizontally on a knife edge placed at $55 c m$ mark. When a mass of $25 g$ is supported from one end, then the mass of the scale is :

What torque will increase angular velocity of a solid disc of mass $16 k g$ and diameter $1 m$ from zero to $2$ rpm in $8 s$ ?