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I did check some answers on similiar questions here, but I wasn't satisfied with the answer. When we are sitting (or if there is any body) on merry - go - round there should be some net force which acts radially in to keep us rotating relatively to ground. In this case it is said that static friction is this force since we are not moving with respect to merry - go - around. We are looking at things from inertial reference frame (no inertial forces). As far as I know static friction opposes motion when some outer force tries to move it from relatively to some other surface (to move us relatively to merry - go - round). So, there should be some force acting radially out to cause static friction acting radially in, but even in that case sum of forces acting in radial direction is zero since static friction balances outer force trying to move us. If so, there is no net force acting radially in to allow rotation. Where have I gone wrong?

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Friction is the centripetal force in this scenario and aside from gravity it is the only force acting on the person on the Merry-go-round.

In an inertial reference frame (say you're standing on the ground looking at the Merry-go-round) and you are staring at person B on the Merry-go-round, you will see Person B moving in a circular motion. From Newton's first law, we know objects with no net force acting on them travel at a constant speed in a straight line which indicates that there is a net force on Person B (static friction acting towards the centre).

static friction opposes motion when some outer force tries to move it...

No, it is truer to say "Static friction always opposes relative motion at the point of contact". This does not necessitate the existence of another force. The centripetal force (static friction) is keeping person B constrained to a circle but is also giving them radial speed. If the force of static friction stopped, they would fly off in a tangential direction.

Person B "wants" to move in a tangential direction to the circular path because their body's inertia "wants" to keep them moving in a straight line, but as the Merry-go-round is rotating, friction will stop them moving in a straight line and act inwards. So in an inertial frame, it isn't a case of force vs force but force vs inertia.

What may confuse you is your interpretation of Newton's 3rd Law. While the force of friction acts on person B, you may think that this law states that they are constrained to the circle because of another force. But this law states that each force acts on a different object. The force of friction acts from the Merry-go-round on Person B but Person B exerts an equal and opposite force in the Merry-go-round itself, so it has nothing to do with person B being constrained to a circle. There is still a net force acting inward, but it has to do with keeping them in that circle. If you increased the centripetal force, the circle would decrease in radius.

If the forces were balanced in the inertial frame, then the net force would be $0$ and according to Newton's first law, Person B would travel in a straight line but they move in a circle.

In the rotating reference frame of the merry-go-round, we don't see Person B move in a circular motion as the reference frame moves with them, but we know the centripetal force is acting inwards so to explain the fact that the net force appears to be $0$ in the rotating reference frame, we "invent" a force (the centrifugal force). This force doesn't exist, in reality it's just the inertia which is why it's often called an inertial force.

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  • $\begingroup$ Good answer, thank you. Yes, friction opposes relative motion and doesn't neccesarily require other force in our scenario static friction opposes inertia. What I am not sure is that due to inertia ,person B would move in the direction of its velocity which is always TANGENT to its circular path. So, friction should act in opposite direction of tangential velocity of person B relative to the ground. If so, static friction doesn't have radial component since it has only opposite orientation of tangential velocity vector. This is what I don't understand. $\endgroup$ Apr 1 at 9:17
  • $\begingroup$ "Static friction always opposes relative motion at the point of contact." The relative motion between the merry-go-round and a person B sitting on it will be directed outwards from the centre. If Person B is on the Merry-go-round and it is spinning counter-clockwise, the point of the MGR under them will move to the left but they want to continue forward so the relative motion is a straight line pointing away from the centre. Static friction therefore points towards the centre. $\endgroup$
    – Aidan
    Apr 1 at 12:57
  • $\begingroup$ Here's a few links which may help: link1 and link2 $\endgroup$
    – Aidan
    Apr 1 at 12:58
  • $\begingroup$ Ohh, yes yes. Thanks, Aidan. What is important to understand this is that if you want to know how static friction acts you need to first consider the situation in which it doesn't act since than inertia will determine what happens. Inertia will determine relative velocity of person B with respect to MGR and than you know in what direction does static friction act since it opposes relative motion. I've sketched it all on paper and concluded that relative velocity of person B with respect to MGR has a component radially out, so static friction on B must act radially in. $\endgroup$ Apr 1 at 17:42
  • $\begingroup$ Correct. The most challenging aspect is picturing what's happening sometimes. Hope this has helped you. $\endgroup$
    – Aidan
    Apr 1 at 17:57
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With the merry-go-round rotating at constant angular velocity, for an object fixed at the edge of the merry-go-round, in the inertial frame (observer on the ground) the object is accelerating due to the change in its velocity vector even though its speed is constant (acceleration is a vector). The force that causes the acceleration is radially inward and is called a centripetal (center seeking) force; in this case the centripetal force is due to friction.

Viewed from a non-inertial reference frame fixed to the object, in that frame the object is at rest and the centripetal force inward due to friction is countered by a centrifugal (center fleeing) force outward. A centrifugal force only appears in a non-inertial reference frame. Other forces can also appear in a non-inertial reference frame, such as the Coriolis force that appears if the object is moving in the non-inertial frame. The forces that only appear in a non-inertial reference frame are called fictitious forces because they are not true forces in the inertial frame, but only appear in the non-inertial frame due to the acceleration of the non-inertial frame.

See a good physics text such as one by Halliday and Resnick for a basic discussion of centripetal and centrifugal forces. For more details on non-inertial reference frames and the fictitious forces that appear therein, see a good physics mechanics text such as Mechanics by Symon.

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Short and simple: friction opposes relative motion between surfaces. You do not need an opposing force for there to be friction.

As a simpler example, consider the case of a block stacked on another block. You gently push on only the bottom block so that the system starts moving as a whole. The static friction acting on the top block is the only force acting on the top block; no additional force is needed to act on the top block for the friction to occur.

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  • $\begingroup$ Yes, that is true. Friction opposes relative motion regardless of force. What I am not sure in my example is that due to inertia body tends to continue moving in a direction of tangential velocity which is always tangential to its circular path. Merry go round doesn't move relatively to body, so in what direction is the tendency for body to move relatively to merry go round? It should have some tendency to move radially out for friction to have component radially in, but how can this be since velocity is always tangential to our path and merry go round doesn't move relatively to us? $\endgroup$ Apr 1 at 8:46
  • $\begingroup$ Yes, the only external force acting on the top block is the friction force applied to it by the bottom block, just like the only external force acting on the body on the merry-go-round is the static friction force applied to it by the merry-go-round. But don't you agree that between the top and bottom blocks you have equal and opposite friction forces per N3, only they are internal forces? And doesn't the same apply to the merry-go-round? This was my intended point. $\endgroup$
    – Bob D
    Apr 1 at 14:38
  • $\begingroup$ Taking the merry-go-round a step further, it exerts an outward force on the ground and the ground exerts an equal and opposite force on the merry-go-round for a net force of zero on the merry-go-round. $\endgroup$
    – Bob D
    Apr 1 at 14:39
  • $\begingroup$ @BobD Yes, you are right, but I was only focusing on the top block, just like how the OP is only focusing on the object on the merry-go-round. The OP had the confusion that an opposing force has to be acting on the object, and that is the point I was trying to address. $\endgroup$ Apr 1 at 15:41
  • $\begingroup$ Yes, but it didn't answer my question which is why does static friction have radial component in the case of merry go round given the points I gave in my previous question? $\endgroup$ Apr 1 at 16:05
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So, there should be some force acting radially out to cause static friction acting radially in, but even in that case sum of forces acting in radial direction is zero since static friction balances outer force trying to move us. If so, there is no net force acting radially in to allow rotation. Where have I gone wrong?

There is a force acting radially out. But as can be seen in the free body diagrams below of an object and the merry-go-round, it is the friction force the object exerts on the merry-go-round in opposition to the friction force (centripetal force) that the merry-go-round exerts on the object, as an action-reaction pair at the surface per Newton's third law. These equal and opposite forces are the reason there is no relative motion between the block and merry-go-round.

However, as shown in the FBD of the block, the friction force the merry-go-round exerts on the object is the only external horizontal force acting on the object and is analogous to the example given in the answer by @BioPhycisist. It is this net force on the object that causes it to accelerate radially inward giving it circular motion.

Hope this helps.

enter image description here

enter image description here

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The friction force on an object which is sitting on a rotating turntable acts toward the center of rotation and provides the required centripetal acceleration.

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  • $\begingroup$ I think the OP understands that. I think the OP wants to know the nature of the force that the static friction force opposes. $\endgroup$
    – Bob D
    Mar 31 at 14:37
  • $\begingroup$ Exactly, Bob D. How does static friction provide centripetal force given everything I said in the text? $\endgroup$ Mar 31 at 17:20
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    $\begingroup$ The friction is not opposing a force. It is causing an acceleration. If you want to work with a rotating (non-inertial) frame of reference, (in which the object is at rest) then you can talk about a (pseudo) centrifugal force. $\endgroup$
    – R.W. Bird
    Mar 31 at 19:00
  • $\begingroup$ @R.W.Bird What is the equal and opposite force to the centripetal (static friction) force, Per Newton's third law? $\endgroup$
    – Bob D
    Mar 31 at 22:02
  • $\begingroup$ @BobD The frictional force by the object on the merry-go-round. It tries to pull merry-go-round mass along a path tangent to the velocity of the object, which would be a $d\vec{v}$ directed radially. $\endgroup$
    – Bill N
    Apr 1 at 0:51

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