Click to Chat

0120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: R

There are no items in this cart.
Continue Shopping
Get instant 20% OFF on Online Material.
coupon code: MOB20 | View Course list

• Complete Physics Course - Class 11
• OFFERED PRICE: R 2,800
• View Details
Get extra R 700 off
USE CODE: SSPD25

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


7 years ago

Share

										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.


7 years ago

# Other Related Questions on Mechanics

Please solve the attached questions with steps. Thanks in advance. Can you please guide me how to solve projectile motion problems and circular motion problem accurately

First of all the answer to the above question is option d and not option c as stated by komali. -.-‘ Now, For projectile motion problems Always resolve motion in two directions , x and y...

 Manas Shukla 2 days ago

the answer is 50m/s and it is surely correct no need to worry about it and proceed please and steps are as follows

 komali 3 days ago
the potential energy U(x) of a particle moving along x-axis is given by U(x) = ax-bx2 .find the equillibrium position of the particle ?

At eqm. position the net Force would be zero. As, F = – dU/dx => -(a -2bx) = 0 a = 2bx x = a/2b would be the eqm. position.

 Vikas TU 7 days ago
A uniform disc of radius R is taken and out of it a disc of diameter R/2 is cut off from end. The center of mass of remaining part will be?

Let M be the mass of the full disc , M’ be the mass of the partial disc and m be the mass of the smaller removed disc. If p is the density of the disc materials Then Now Lets assume the...

 Manas Shukla 3 days ago
WHAT IS A OVER DAMPED OSCILLATIONS...........................?

Dear Gangadhar, Over damped oscillations means the amplitude is decreases in gradually eith making some cycles.

 SAI SARDAR 7 months ago

Over Damped : Over damping of a damped oscillator will cause it to approach ... Under Damping: The system oscillates (at reduced frequency compared to the underdamped oscillator)

 Prabhakar ch 7 months ago
WHAT ARE THE USES OF MAGNETS ?WHERE THEY ARE USED EXTENSIVELY?

Hello Jyothi The attraction a magnet produces is called a "magnetic field." ... Other kinds of magnets are even more powerful, strong enough to pick up ... Usage Examples.

my dear friend it was simple that the uses of magnets are very simple was that to attract something which is related to iron as well as some metals.there are many uses of magnets in our...

 NAVEENKUMAR NAGIPOGU 9 months ago

jyotheeswar the attraction a magnet products is called a magnetic field and other kinds of magnets are even more powerful strong enough to pick up and more

waves are required for us in what way or not

to know aboute energy in distrubence

 Gavvala Ganesh one year ago

 L GOUSE BASHA one year ago

 pa1 one year ago
View all Questions »

• Complete Physics Course - Class 12
• OFFERED PRICE: R 2,600
• View Details
Get extra R 650 off
USE CODE: SSPD25

• Complete Physics Course - Class 11
• OFFERED PRICE: R 2,800
• View Details

Get extra R 700 off
USE CODE: SSPD25

More Questions On Mechanics