# How does centrifugal force count on the ball-rope problem?

I'm whirling a ball tying it by a rope. Then at any instant the centrifugal force (outward force) equalizes the centripetal force (inward force) . Then both should cancel each other and there should be only tension of the rope to balance the weight of the ball. Then why do we count the centripetal force as resultant force. As an example in the mean position $T=mg + m\frac{v^2}{R}$ why do we also consider the second term?

• The idea is not to think of centripetal force as its own separate force, but a condition that is required to keep an object in uniform circular motion. In fact you should start these problems with Newton's second written as $\sum F = \frac{mv^2}{R}$ where the term on the left are the forces exerted on the ball, whose net effect must satisfy the centripetal force cobdition – Triatticus Apr 28 '18 at 1:40
• @Triatticus if centrifugal force equals the centripetal force. Then any force should not be on the particle and shouldn't it fall? – Nobody recognizeable Apr 28 '18 at 2:39

Suppose, you are analysing the situation from the your frame of reference. Then, the net radial inward force on the ball is, $$T - mg \sin \theta$$, while the inward radial acceleration is $$\frac{v^2}{r}$$. Applying Newton's second law, $$T - mg \sin \theta$$ $$=\frac{mv^2}{r}$$.

Now if you analyse the motion from the frame of the ball, there is no radial acceleration. So, the net radial force must be zero. The inward radial force is, again, $$T - mg \sin \theta$$, while there is the centrifugal force (pseudo force) along the outward radial direction having magnitude $$\frac{mv^2}{r}$$. So, we get, $$\frac{mv^2}{r}-(T-mg \sin \theta) =0$$. Thus, the equations are same in both of the frames.

From the reference frame of you whirling the ball, there is no centrifugal force. The centrifugal force is a fictitious force, only showing up in the reference frame of the ball. This is because the ball's reference frame is not inertial (it is traveling in a circle, and therefore is constantly accelerating). Noninertial frames gives rise to fictitious forces like the centrifugal and Coriolis forces.

The tension is the centripetal force (it is what keeps the ball moving in a circle). So the only forces acting on the ball, from your reference frame, are the centripetal force (tension) and gravity (neglecting air resistance). There is no centrifugal force, in your frame, to cancel the centripetal. If there was a force cancelling the centripetal force, there wouldn't be circular motion.

• If that is true then we should write $T=mg$ always. – Nobody recognizeable Apr 28 '18 at 0:27
• What is $T$? Tension? – roshoka Apr 28 '18 at 0:28
• tension is a force which takes accelerates the ball towards the center. – Nobody recognizeable Apr 28 '18 at 0:32
• $T$ does not always equal $mg$. Think about the direction of each force. Gravity is always downward, but tension is always pointed toward the center of the circle that the ball is moving in. So gravity and tension will not always be in the same direction. So you can't say $\vec{T}=m\vec{g}$ always. – roshoka Apr 28 '18 at 0:33
• Are you spinning this ball horizontally or vertically? – roshoka Apr 28 '18 at 0:42