Physics System of Particles and Rotational Motion
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New answer posted
a month agoContributor-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 agoContributor-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.
New answer posted
a month agoContributor-Level 10

Let the gravitational field is zero at a distance from the mass .
or
Gravitational potential at
New answer posted
a month agoContributor-Level 10
Radius of gyration of a solid surface,
Radius of gyration of a hollow surface,
New answer posted
a month agoContributor-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 agoNew answer posted
a month agoContributor-Level 10
The x and y coordinates of the center of mass are given by:
x? = y? = 4a / 3π
New answer posted
a month agoContributor-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 agoContributor-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 agoContributor-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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