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Physics Part-I

System of Particles and Rotational Motion

A non-uniform bar of weight W is suspended at rest by two stri

Physics Part-I

Class 11

NCERT

Chapter 1: Physical World

16 questions

Chapter 2: Units and Measurements

51 questions

Chapter 3: Motion in Straight Line

50 questions

Chapter 4: Motion in a Plane

43 questions

Chapter 5: Laws of motion

52 questions

Chapter 6: Work, Energy and Power

41 questions

Chapter 7: System of Particles and Rotational Motion

45 questions

Two particles, each of mass

m

and speed

v

, travel in opposite directions along parallel lines separated by a distance

d

. Show that the vector angular momentum of the two particle system is the same whatever be the point about which the angular momentum is taken.

A non-uniform bar of weight

W

is suspended at rest by two strings of negligible weight as shown in fig. The angles made by the strings with the vertical are

36. 9^{o}

and

53. 1^{o}

respectively. The bar is

2 m

long. Calculate the distance d of the center of gravity of the bar from its left end.

A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Detere the force exerted by the level ground on each front wheel and each back wheel.

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Chapter 8: Gravitation

33 questions

Question

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Solving time: 3 mins

A non-uniform bar of weight $W$ is suspended at rest by two strings of negligible weight as shown in fig. The angles made by the strings with the vertical are $36. 9^{o}$ and $53. 1^{o}$ respectively. The bar is $2 m$ long. Calculate the distance d of the center of gravity of the bar from its left end.

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Updated on: Dec 17, 2023

Book:

Physics Part-I (NCERT)

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Liam Discussed

A non-uniform bar of weight

W

is suspended at rest by two strings of negligible weight as shown in fig. The angles made by the strings with the vertical are

36. 9^{o}

and

53. 1^{o}

respectively. The bar is

2 m

long. Calculate the distance d of the center of gravity of the bar from its left end.

13 mins ago

Discuss this question LIVE

13 mins ago

Text solutionVerified

The free body diagram of the bar is shown in the following figure.

Length of the bar is ,l=2m

T_{1}, T_{2}

be the tensions produced in the left and right strings respectively.

At translational equilibrium, we have,

T_{1} s in 36. 9^{\circ} = T_{2} s in 53. 1^{\circ}

⟹ \frac{T _{1}}{T _{2}} = \frac{4}{3}

For rotational equilibrium, on taking the torque about the centre of gravity, we have:

T_{1} (cos 36. 9^{\circ}) \times d = T_{2} cos 53. 1^{\circ} (2 - d)

Using both equations,

d = 0.72 m

Hence, the centre of gravity of the given bar lies 0.72 m from its left end.

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Practice more questions from Physics Part-I (NCERT)

Two particles, each of mass

m

and speed

v

, travel in opposite directions along parallel lines separated by a distance

d

. Show that the vector angular momentum of the two particle system is the same whatever be the point about which the angular momentum is taken.

A non-uniform bar of weight

W

is suspended at rest by two strings of negligible weight as shown in fig. The angles made by the strings with the vertical are

36. 9^{o}

and

53. 1^{o}

respectively. The bar is

2 m

long. Calculate the distance d of the center of gravity of the bar from its left end.

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Question Text	A non-uniform bar of weight $W$ is suspended at rest by two strings of negligible weight as shown in fig. The angles made by the strings with the vertical are $36. 9^{o}$ and $53. 1^{o}$ respectively. The bar is $2 m$ long. Calculate the distance d of the center of gravity of the bar from its left end.
Updated On	Dec 17, 2023
Topic	System of Particles and Rotational Motion
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 5
Upvotes	592
Avg. Video Duration	12 min

Text solutionVerified

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A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in fig. The angles made by the strings with the vertical are 36.9o and 53.1o respectively. The bar is 2 m long. Calculate the distance d of the center of gravity of the bar from its left end.

Text solutionVerified

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