Question

Speed of an object in circular motion reduces at a rate of 3m/s every second which is equal to its centripetal acceleration. What is the direction of resultant acceleration with X- axis if centre of circular path is at the origin.​

Answer 1

Answer:

A centripetal force is a net force that acts on an object to keep it moving along a circular path.

In our article on centripetal acceleration, we learned that any object traveling along a circular path of radius rrr with velocity vvv experiences an acceleration directed toward the center of its path,

a = \frac{v^2}{r}a=rv2a, equals, start fraction, v, squared, divided by, r, end fraction.

However, we should discuss how the object came to be moving along the circular path in the first place. Newton’s 1ˢᵗ law tells us that an object will continue moving along a straight path unless acted on by an external force. The external force here is the centripetal force.

It is important to understand that the centripetal force is not a fundamental force, but just a label given to the net force which causes an object to move in a circular path. The tension force in the string of a swinging tethered ball and the gravitational force keeping a satellite in orbit are both examples of centripetal forces. Multiple individual forces can even be involved as long as they add up (by vector addition) to give a net force towards the center of the circular path.

Starting with Newton's 2ⁿᵈ law :

a = \frac{F}{m}a=mFa, equals, start fraction, F, divided by, m, end fraction

and then equating this to the centripetal acceleration,

\frac{v^2}{r} = \frac{F}{m}rv2=mFstart fraction, v, squared, divided by, r, end fraction, equals, start fraction, F, divided by, m, end fraction

We can show that the centripetal force F_cFcF, start subscript, c, end subscript has magnitude

F_c = \frac{mv^2}{r}Fc=rmv2F, start subscript, c, end subscript, equals, start fraction, m, v, squared, divided by, r, end fraction

and is always directed towards the center of the circular path. Equivalently, if \omegaωomega is the angular velocity then because v=r\omegav=rωv, equals, r, omega,

F_c = m r \omega^2Fc=mrω2