In the system shown, a force of \$100\$ N is applied at the end shown. What is the magnitude \$\tau \$ (in N-m) of the torque which was produced on the drum for starting motion? (Given that the coefficient of static friction is \$0.1\$)

For the motion to start, the 'moment-couple' on the drum must be equal to the product of frictional force of the rod against the drum and the drum radius. Since the rod is in equilibrium, we can take torque about the fulcrum to find the normal force of the drum on the rod (Normal force N acts at centre of rod)\$\sum \tau _{fulcrum}=0\$or\$ -100(2+2)+N(2)=0\$or\$N=200\ N\$ Friction force\$f=\mu N\$ \$= (0.1)(200) = 20 N \$\$\tau =\$ Torque produced by couple\$=\$\$(20)1\$or \$\tau =20\ Nm\$

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>>Class 11
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>>Systems of Particles and Rotational Motion
>> $Hard Questions$

Systems Of Particles And Rotational Motion

Q: Four thin metal rods, each of mass \$M\$ and length \$L\$, are welded to form a square. The moment of inertia of the composite structure about a line which bisects any two opposite rods is:

The MI of the rod AB(\$I_1\$ )and CD(\$I_2\$ ) about the axis PQ( as shown in the figure ) are same.\$ I_1=I_2=\dfrac{ML^2}{12}\$The MI of the rod BC(\$I_3\$ )and AD(\$I_4\$ ) about the axis PQ( as shown in the figure ) are same..\$ I_3=I_4={M}{(\dfrac{L}{2})^2}\$\$ AD \; \left | \right | \:PQ\$ and \$ BC \; \left | \right | \:PQ\$ and all points on AD are at same distance from the axis PQ, all points on BC are at same distance from the axis PQ, \$ \therefore I =I_1 + I_2 +I_3 +I_4= 2 \times \dfrac{ML^2}{12} + 2 \times {M}{(\dfrac{L}{2})^2}=\dfrac{2}{3}ML^2\$

Q: Three identical spheres each of mass m and radius R are placed touching each other so that their centres A, B and C lie on a straight line. The position of their centre of mass from the centre of A is

[B] From Symmetry B is the centre of ABC combined.Position of centre of mass of ABC (combined) fromcentre of \$A = R + R = 2R\$

Q: The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its midpoint and perpendicular to its length is \$0_0\$. Its moment of inertia about an axis passing through one of its ends and perpendicular to its length is :

Let the end points of the rod are designated as A and B.Applying parallel axis theorem,Given, \$I_{cm}=O_o\$\$I_A= I_{cm} + M(\frac{L}{2})^2= O_o + M\frac{L^2}{4}\$

Q: An automobile is started from rest with one of its doors initially at right angles. lf the hinges of the door are toward the front of the car, the door will shut as the automobile picks up speed. The acceleration a is constant and the centre of mass is at a distance d from the hinges. The time \$T\$ needed for the doors to close is given by: \$\left [ Given:\displaystyle\int_{0}^{\frac{\pi }{2}}\dfrac{d\theta }{\sqrt{sin\theta }}=N \right ]\$(Consider the door as a square plate)

Equation of motion is \$I\alpha =Ma(dcos\theta )\$Now moment of inertia of the door about the hinges is given as\$I=\dfrac{Md^{2}}{3}+Md^{2}\$\$=M\left ( \dfrac{4}{3}d^{2} \right )\$Hence,\$\dfrac{4}{3}Md^{2}\alpha =Ma dcos\theta \$\$\Rightarrow \dfrac{4}{3}d\times\omega \dfrac{d\omega }{d\theta }=acos\theta \left ( \because \alpha =\omega \dfrac{d\omega }{d\theta } \right )\$\$\Rightarrow \dfrac{4}{3}d\int \omega d\omega =\int acos\theta d\theta \$\$\Rightarrow \dfrac{4}{3}d\dfrac{\omega ^{2}}{2}=asin\theta \$\$\omega ^{2}=\dfrac{3a}{2d}sin\theta +C\$where, C\$=\$constant of integrationAt \$\theta =0\$ (initially), we have \$\theta =0\$\$\Rightarrow C=0\$\$\Rightarrow \omega^{2}=\dfrac{3a}{2d}sin\theta \$\$\Rightarrow \omega =\sqrt{\dfrac{3a}{2d}sin\theta }\$\$\dfrac{d\theta }{\sqrt{sin\theta }}=\sqrt{\dfrac{3a}{2d}}dt\$ and integrating between \$t=0\$ and \$t=T\$(corresponding to \$\theta =0\$ and \$\theta =\dfrac{\pi }{2}\$)\$\int_{0}^{\dfrac{\pi }{2}}\dfrac{d\theta }{\sqrt{sin\theta }}=\sqrt{\dfrac{3a}{2d}}.T\$\$\Rightarrow T=\left ( \sqrt{\dfrac{2d}{3a}} \right )N\$\$\left ( since,given:\displaystyle N=\int_{0}^{\dfrac{\pi }{2}}\dfrac{d\theta }{\sqrt{sin\theta }} \right )\$

Q: Three identical thin rods each of mass \$m\$ and length \$L\$ are joined together to form an equilateral triangular frame. The moment of inertia of frame about an axis perpendicular to the plane of frame and passing through a corner is:

Sum of MI of the two rods, at vertex of which the axis passes, is \$2\times mL^2/3=2mL^2/3\$For the third rod, the distance between center of the rod and the axis is \$d=\sqrt3L/2\$Using parallel axis theorem, its MI is \$mL^2/12+3mL^2/4=5mL^2/6\$So total MI of the system is \$2mL^2/3+5mL^2/6=3mL^2/2\$

Q: A thin uniform metallic triangular sheet of mass M has sides \$AB=BC=L\$. Its moment of inertia about the axis AC lying in the plane of the sheet is:

The given triangular sheet has an area equal to half of a square sheet having same side length as the length of the base of the triangle.Let MI of the triangular sheet about AC be I.I is half of the MI of the square plate about an axis passing through the centre O and being in the plane of the plate.Let the MI of the square about an axis XY be \$I_{XY}\$.Due to symmetry\$I_{AB} = I_{CD}\$ AND \$I_{PQ} = I_{RS}\$Let \$I_{square}\$ be the MI a bout an axis passing through the centre of the square and perpendicular to the plane of the square plate.Then using Perpendicular Axis Theorem\$I_{square} = I_{AB} + I_{CD}=I_{PQ} + I_{RS} \$\$ \therefore I_{square} = 2 I_{AB} = I_{PQ} \$But, \$I_{AB}= 2I\$\$ \therefore I = \dfrac{1}{2}I_{AB} = \dfrac{1}{2} *(\dfrac{1}{2}\dfrac{ML^2}{6})\$ ....(1) where M is the mass of the square plate.But the triangular sheet has half the mass of the square plate. Writing I in terms of the mass of the plate.\$ \therefore I =\dfrac{M}{2}*L^2/12=\dfrac{1}{12}M_{triangle}L^2\$

Q: A car (open at the top) of mass 9.75 kg is coasting along a level track at 1.36 m/s, when it begins to rain hard. The raindrops fall vertically with respect to the ground. When it has collected 0.5 kg of rain, the speed of the car is

Initial \$m_{1} = 9.75\$ kgfinal \$m =9.75 + 0.5 = 10.25\$\$m_{1}v_{1} = m_{2}v_{2}\$\$(9.75)(1.36) = (10.25)v_{2}\$\$v_{2} = 1.29\$ m/s

Q: A uniform rod AB of mass m and length \$\ell\$ is in equilibrium with the help of two identical inextensible strings as shown in the figure. Choose the correct option (s)

The point \$A\$ will move in a circular motion after cutting the string\$2\$. But, relative to the point \$A\$, the center of mass will also move in a circular motion, keeping \$A\$ its center. Therefore in the ground frame, the motion of the center of mass will not be circular.The instantaneous velocity of the point \$A\$ will be 0 after cutting the string\$2\$. And it will vary with the acceleration \$a = \dfrac{{{v^2}_{{\text{instant}}}}}{r}\$. Therefore, its acceleration along the string will be \$0\$.The acceleration of the point \$A\$ perpendicular to the string\$1\$ is not zero because of the unbalanced forces along the direction.

Physics
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BITSAT Easy Qs Med Qs Hard Qs

Four thin metal rods, each of mass $M$ and length $L$ , are welded to form a square. The moment of inertia of the composite structure about a line which bisects any two opposite rods is:

\frac{M L ^{2}}{6}

\frac{M L ^{2}}{3}

\frac{M L ^{2}}{2}

\frac{2 M L ^{2}}{3}

Hard

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Three identical spheres each of mass m and radius R are placed touching each other so that their centres A, B and C lie on a straight line. The position of their centre of mass from the centre of A is

\frac{2 R}{3}

\frac{5 R}{3}

\frac{4 R}{3}

Hard

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The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its midpoint and perpendicular to its length is $0_{0}$ . Its moment of inertia about an axis passing through one of its ends and perpendicular to its length is :

0_{0} + \frac{M L ^{2}}{2}

0_{0} + \frac{M L ^{2}}{4}

0_{0} + 2 M L^{2}

0_{0} + M L^{2}

Hard

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An automobile is started from rest with one of its doors initially at right angles. lf the hinges of the door are toward the front of the car, the door will shut as the automobile picks up speed. The acceleration a is constant and the centre of mass is at a distance d from the hinges. The time $T$ needed for the doors to close is given by:
$[G i v e n : \int_{0}^{\frac{π}{2}} \frac{d θ}{s i n θ} = N]$ (Consider the door as a square plate)

T = (\frac{3 d}{2 a}) N

T = (\frac{d}{3 a}) N

T = (\frac{2 d}{a}) N

T = (\frac{2 d}{3 a}) N

Hard

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Three identical thin rods each of mass $m$ and length $L$ are joined together to form an equilateral triangular frame. The moment of inertia of frame about an axis perpendicular to the plane of frame and passing through a corner is:

\frac{2 m L ^{2}}{3}

\frac{3 m L ^{2}}{2}

\frac{4 m L ^{2}}{3}

\frac{3 m L ^{2}}{4}

Hard

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A uniform sheet each of thickness 10 units is
cut into the shape as shown. Compute then x
and y-coordinates of the centre of mass of the
piece from point A.

17.5, 22.5

22.5, 17.5

10.5, 22.5

190/7, 125/7

Hard

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A thin uniform metallic triangular sheet of mass M has sides $A B = B C = L$ . Its moment of inertia about the axis AC lying in the plane of the sheet is:

\frac{M L ^{2}}{1 2}

\frac{M L ^{2}}{6}

\frac{M L ^{2}}{3}

\frac{2 M L ^{2}}{3}

Hard

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In the system shown, a force of $100$ N is applied at the end shown. What is the magnitude $τ$ (in N-m) of the torque which was produced on the drum for starting motion? (Given that the coefficient of static friction is $0.1$ )

10

5

20

25

Hard

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A car (open at the top) of mass 9.75 kg is coasting along a level track at 1.36 m/s, when it begins to rain hard. The raindrops fall vertically with respect to the ground. When it has collected 0.5 kg of rain, the speed of the car is

0.68 m/s

1.29 m/s

2.48 m/s

9.8 m/s

Hard

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A uniform rod AB of mass m and length $ℓ$ is in equilibrium with the help of two identical inextensible strings as shown in the figure. Choose the correct option (s)

After cutting the string (2) motion of centre of mass of the rod is circular

After cutting the string (2) motion of centre of mass of the rod is not circular

Just after cutting the string (2) the acceleration of point A along the string (1) is zero.

Just after cutting the string (2) the acceleration of point A perpendicular to the string (1) is not zero.

Hard

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Systems Of Particles And Rotational Motion

Physics 5225 Views

Four thin metal rods, each of mass M and length L, are welded to form a square. The moment of inertia of the composite structure about a line which bisects any two opposite rods is:

Three identical spheres each of mass m and radius R are placed touching each other so that their centres A, B and C lie on a straight line. The position of their centre of mass from the centre of A is

The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its midpoint and perpendicular to its length is 00​. Its moment of inertia about an axis passing through one of its ends and perpendicular to its length is :

Three identical thin rods each of mass m and length L are joined together to form an equilateral triangular frame. The moment of inertia of frame about an axis perpendicular to the plane of frame and passing through a corner is:

A uniform sheet each of thickness 10 units is cut into the shape as shown. Compute then x and y-coordinates of the centre of mass of the piece from point A.

A thin uniform metallic triangular sheet of mass M has sides AB=BC=L. Its moment of inertia about the axis AC lying in the plane of the sheet is:

In the system shown, a force of 100 N is applied at the end shown. What is the magnitude τ (in N-m) of the torque which was produced on the drum for starting motion? (Given that the coefficient of static friction is 0.1)

A car (open at the top) of mass 9.75 kg is coasting along a level track at 1.36 m/s, when it begins to rain hard. The raindrops fall vertically with respect to the ground. When it has collected 0.5 kg of rain, the speed of the car is

A uniform rod AB of mass m and length ℓ is in equilibrium with the help of two identical inextensible strings as shown in the figure. Choose the correct option (s)

Physics
5225 Views

Four thin metal rods, each of mass $M$ and length $L$ , are welded to form a square. The moment of inertia of the composite structure about a line which bisects any two opposite rods is:

The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its midpoint and perpendicular to its length is $0_{0}$ . Its moment of inertia about an axis passing through one of its ends and perpendicular to its length is :

Three identical thin rods each of mass $m$ and length $L$ are joined together to form an equilateral triangular frame. The moment of inertia of frame about an axis perpendicular to the plane of frame and passing through a corner is:

A uniform sheet each of thickness 10 units is
cut into the shape as shown. Compute then x
and y-coordinates of the centre of mass of the
piece from point A.

A thin uniform metallic triangular sheet of mass M has sides $A B = B C = L$ . Its moment of inertia about the axis AC lying in the plane of the sheet is:

In the system shown, a force of $100$ N is applied at the end shown. What is the magnitude $τ$ (in N-m) of the torque which was produced on the drum for starting motion? (Given that the coefficient of static friction is $0.1$ )

A uniform rod AB of mass m and length $ℓ$ is in equilibrium with the help of two identical inextensible strings as shown in the figure. Choose the correct option (s)