Physics System of Particles and Rotational Motion

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a month ago

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A
Aadit Singh Uppal

Contributor-Level 10

Even if the torque is same, the rotation of a body can be slow or fast depending on the distribution of mass along the surface area. If the mass of a body is more on the axis area, it's speed of rotation will be faster as compared to the body whose mass is accumulated away from the axis due to which the body will resist the angular acceleration.

New answer posted

a month ago

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A
Aadit Singh Uppal

Contributor-Level 10

In simple words, a push or pull applied on a body is known as a force. Force is a concept used in terms of linear motion. On the other hand, the force which causes a body to spin around an axis instead of moving from one point to aother is called torque. Torque is a theory which comes into account while studying rotational motion and is considered as the rotational analogue of force.

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a month ago

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alok kumar singh

Contributor-Level 10

Let the gravitational field is zero at a distance x from the mass m .

G m x 2 = G 9 m ( R - x ) 2

R - x = 3 x or x = R 4

Gravitational potential at  R 4

= - G m R 4 - G 9 m 3 R 4

= - 4 G m R - 12 G m R

= - 16 G m R

 

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a month ago

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A
alok kumar singh

Contributor-Level 10

Radius of gyration of a solid surface,

K S = 2 5 R

Radius of gyration of a hollow surface,

K H = 2 3 R

K S K H = 3 5 = 3 5

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

μ_min = (I tanθ)/ (I + mR²)
= [ (mR²/2)tanθ] / [mR²/2 + mR²] = (tanθ)/3 = (tan 60°)/3 = √3/3 = 1.732/3 = 0.5773

Since given coefficient of static friction is less than μ_min, so body will perform rolling with slipping and kinetic friction will act
F? = μN = μmg cosθ = (0.4) * mg cos 60° = mg/5

New question posted

a month ago

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New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

The x and y coordinates of the center of mass are given by:
x? = y? = 4a / 3π

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a month ago

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R
Raj Pandey

Contributor-Level 9

Using conservation of Angular momentum along axis of rotation, we can write
Mr²ω = (Mr² + 2mr²)ω? ⇒ ω? = Mω / (M + 2m)

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

The number of revolutions can be found using the rotational kinematic equation for angular displacement (θ):
θ = (ω_initial + ω_final)/2 * t
Number of revolutions = θ / 2π
Number of revolutions = [ (ω_final + ω_initial) * t] / (2 * 2π)

Based on the numerical values provided in the document, the calculation is:
Number of revolution = [ (2π * 3360/60 + 0) * t] / (2 * 2π) . with further calculation yielding the result:
Number of revolution = 728

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

The equations for an object rolling down an inclined plane without slipping are:

·       Force equation: mg sinθ - f_s = ma

·       Torque equation: f_s R = Iα

Since a = αR, we can write f_s = Iα/R = Ia/R².
Substituting this into the force equation:
mg sinθ - Ia/R² = ma
mg sinθ = a (m + I/R²)
a = (mg sinθ) / (m + I/R²)

The time taken to travel a distance S is given by S = ½ at², which means t ∝ 1/√a. Therefore, the object with the largest acceleration (a) will arrive first.

The problem is analyzed for different bodies:

·       Ring: I =

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