Question

an object of mass 20 gram is executing uniform circular motion on a circular path of radius 2 metre with the time period 1 of 1 by 2 second find centripetal force acting on it

Answer 1

solution attached........

Answer 2

Answer:

F_c\ =\ 6.31\ N

Explanation:

Given,

• Mass of the object = m = 20 = 0.020 kg
• Radius of the circular motion = R = 2 m
• Time period of the object = $T = \dfrac{1}{2}$

As we know that the time period of an object moving in a circular motion is

$T\ =\ \dfrac{2\pi R}{v}$

Where R is the radius of the circular path and v is the average velocity of the object through out the motion, but the object is moving uniformly through out the motion, hence the average velocity of the object is equal to the velocity of the object,

Let v be the velocity of the object

From the time period,

$T\ =\ \dfrac{2\pi R}{v}\\\Rightarrow v\ =\ \dfrac{2\pi R}{T}\\\Rightarrow v\ =\ \dfrac{2\times 3.14\times 2}{\dfrac{1}{2}}\\\Rightarrow v\ =\ 25.12\ m/s.$

Now the centripetal force be $F_c$ acting on the object.

$\therefore F_c\ =\ \dfrac{mv^2}{R}\\\Rightarrow F_c\ =\ \dfrac{0.020\times 25.12^2}{2}\\\Rightarrow F_c\ =\ 6.31\ N$