Question

object of mass 2kg rotates at constant speed in a horizontal circle of radius 5 metre the time for one complete revolution of 3 seconds what is the magnitude of resultant force acting on the object​

Answer 1

Answer:

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Explanation:

m = 5 kg

r = 5 m

F = mω²r

⇒ ω = 2π/T = 2π/3

On solving we get F = 40π²/9

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Answer 2

Answer:

The magnitude of resultant force acting on the object​ is $\frac{40\pi^{2} }{9}\ Newton$.

Explanation:

Given:

Object of mass 2 kg rotates at constant speed in a horizontal circle of radius 5 meter.

The time for one complete revolution of 3 seconds.

To Find:

The magnitude of resultant force acting on the object​.

Formula Used:

The force acting on an object in rotational motion can be given as:

$F =mr w^{2}$  ----------- formula no. 01

Where.

m= mass of an object

r = radius of circle

w= angular velocity

$w=\frac{2\pi }{T}$  ------  formula no. 02

Where,

w= angular velocity

T= time period

Solution:

As given,object of mass 2 kg rotates at constant speed in a horizontal circle of radius 5 meter.

m=2 kg     and r= 5 meter

As given,the time for one complete revolution of 3 seconds.

T= 3 seconds

Applying formula no.02.

Angular velocity, $w=\frac{2\pi }{3}$

Applying formula no.01.

Resultant Force ,​$F=2\times 5\times ( \frac{2\pi }{3} )^{2}$

   $=10\times ( \frac{4\pi^{2} }{9} )$

  $= \frac{40\pi^{2} }{9}$ Newton.

Thus, the magnitude of resultant force acting on the object​ is $\frac{40\pi^{2} }{9}\ Newton$.

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