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rewona [7]
1 year ago
10

If you see an object rotating, is there necessarily a net torque acting on it?

Physics
1 answer:
mina [271]1 year ago
7 0

There need not necessarily be net torque acting on a rotating object.

<h3>What is angular acceleration?</h3>
  • Angular acceleration is the term used to describe the temporal pace at which angular velocity varies. Radians per second is the accepted unit of measurement. Consequently, = d d t. Angular acceleration is also known as rotational acceleration.
  • Angular acceleration can be calculated by dividing the angular velocity by the acceleration duration. (t). Use pi instead, or the drive speed (n) divided by the acceleration time (t) multiplied by 30. The usual SI unit for the rotational acceleration that results from this equation is radians per second squared (Rad/sec2).

Is there always a net torque acting on an object when you perceive it rotating?

  • Torque, force, and angular acceleration are the ideas applied to this problem's solution.
  • Use the notion of force to first determine whether there is net torque operating on a rotating body.
  • Then, using the idea of angular acceleration, determine whether there is net torque operating on a rotating body.
  • Torque is the measure of the amount of force that causes a body to accelerate in one direction. It is often referred to as the rotational force of a body.

Step: 1

  • The torque of a spinning body can be zero if an object's total amount of forces acting on it is zero or if the object is in rotational equilibrium.
  • A vector quantity is a torque. There won't be any net torque acting on the body if the forces causing the rotation are equal and opposing.
  • Using the idea of angular acceleration, determine whether there is net torque operating on a rotating body.
  • The body's angular acceleration will be 0 if it is rotating with constant angular speed. The net torque is 0 for a body with 0 angular acceleration.

There need not necessarily be net torque acting on a rotating object.

To learn more about angular acceleration, refer to:

brainly.com/question/20912191

#SPJ4

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Explanation:

Formula which holds true for a leans with radii R_{1} and R_{2} and index refraction n is given as follows.

          \frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

Since, the lens is immersed in liquid with index of refraction n_{1}. Therefore, focal length obeys the following.  

            \frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]  

             \frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

and,       \frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}

or,          f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}

              f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}

                          = 32.4 cm

Using thin lens equation, we will find the focal length as follows.

             \frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}

Hence, image distance can be calculated as follows.

       \frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}

              s_{i} = \frac{fs_{o}}{s_{o} - f}

             s_{i} = \frac{32.4 \times 100}{100 - 32.4}

                       = 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

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