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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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A 1300 kg steel beam is supported by two ropes. (Figure
Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

  • the net horizontal force acting on the beam is

R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0

where R_1,R_2 are the magnitudes of the tensions in ropes 1 and 2, respectively;

  • the net vertical force acting on the beam is

R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0

where m=1300\,\rm kg and g=9.8\frac{\rm m}{\mathrm s^2}.

Eliminating R_2, we have

\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)

R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2

R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2

-R_1 \sin(50^\circ) = -\dfrac{mg}2

R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}

Solve for R_2.

\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0

\dfrac{R_2}2 = -mg\cot(110^\circ)

R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}

8 0
2 years ago
What is the average speed of an aircraft which travels 600m in 10 seconds ?
matrenka [14]

Answer:

60 meters per second

Explanation:

600/10=60

8 0
3 years ago
Which of the following would add more resistance? Check all that apply.
Blababa [14]

Answer:C. E.

Explanation:

4 0
3 years ago
1. Calculati greutatea unui sac cu 5 kg de cartofi într-o zonă in care acceleratia gravitatională
harkovskaia [24]

Answer:

the weight is 49.1 N

Explanation:

The computation of the weight is shown below:

As we know that

= 5kg of potatoes × gravitational acceleration

= 5kg of potatoes × 9.82 m/s

= 49.1 N

Hence, the weight is 49.1 N

We simply applied the above formula in order to determine the weight

6 0
3 years ago
A horizontal spring with spring constant 200N/m is compressed by 15cm and used to launch a 2kg box across a frictionless horizon
kobusy [5.1K]

Answer:

Explanation:

Given that,

Spring constant k=200N/m

Compression x = 15cm = 0.15m

Attached mass m =2kg

Coefficient of kinetic friction uk= 0.2

The energy in the spring is given as

U =½kx²

U = ½ × 200 × 0.15²

U = 2.25J

Force in the spring is given by Hooke's law

F = ke

F = 200×0.15

F = 30N

The weight of body which is equal to the normal is give as

W = mg

W = 2 × 9.81

W = 19.62N

W = N = 19.62 Newton's 2nd Law

From law of friction,

Fr = uk•N

Fr = 0.2 × 19.62

Fr = 3.924

Using newton second law again

Fnet = F - Fr

Fnet = 30 - 3.924

Fnet = 26.076

Work done by net force is given as

W = Fnet × d

W = 26.076d

Then, the work done by this net force is equal to the energy in the spring

W = U

26.076d = 2.25

d = 2.25/26.076

d = 0.0863m

Which is 8.63cm

So the box will slide 8.63cm before stopping

6 0
3 years ago
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