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can you please explain me the different cases in torque?with figures

 can you please explain me the different cases in torque?with figures

Grade:12

1 Answers

AskiitianExpert Shine
10 Points
12 years ago

Hi

Different cases with the figures:

  • Moment arm formula


Moment arm diagram

A very useful special case, often given as the definition of torque in fields other than physics, is as follows:

|\tau| = (\textrm{moment\ arm}) (\textrm{force}).

The construction of the "moment arm" is shown in the figure below, along with the vectors r and F mentioned above. The problem with this definition is that it does not give the direction of the torque but only the magnitude, and hence it is difficult to use in three-dimensional cases. If the force is perpendicular to the displacement vector r, the moment arm will be equal to the distance to the centre, and torque will be a maximum for the given force. The equation for the magnitude of a torque, arising from a perpendicular force:

|\tau| = (\textrm{distance\ to\ center}) (\textrm{force}).
  • Force at an angle

If a force of magnitude F is at an angle θ from the displacement arm of length r (and within the plane perpendicular to the rotation axis), then from the definition of cross product, the magnitude of the torque arising is:

 
τ = rFsinθ
  • Static equilibrium

For an object to be in static equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a two-dimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically determinate equilibrium problems in two-dimensions, we use three equations.

  • Torque as a function of time


The torque caused by the two opposing forces Fg and -Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess.

Torque is the time-derivative of angular momentum, just as force is the time derivative of linear momentum:

\boldsymbol{\tau} ={\mathrm{d}\mathbf{L} \over \mathrm{d}t} \,\!

where

L is angular momentum.

Angular momentum on a rigid body can be written in terms of its moment of inertia \boldsymbol I \,\! and its angular velocity \boldsymbol{\omega}:

\mathbf{L}=I\,\boldsymbol{\omega} \,\!

so if \boldsymbol I \,\! is constant,

\boldsymbol{\tau}=I{\mathrm{d}\boldsymbol{\omega} \over \mathrm{d}t}=I\boldsymbol{\alpha} \,\!

where α is angular acceleration, a quantity usually measured in radians per second squared.

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