Uniform vertical circular motion

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Homework Help Overview

The problem involves an object of mass m attached to a light rod performing uniform vertical circular motion. The centripetal force is provided by the tension in the rod and the weight of the object. The original poster seeks to understand the net force acting on the mass when it is at a height of r from the bottom of the circular path, specifically through a free body diagram.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the direction of the net force, questioning why it appears to point slightly downwards rather than directly towards the center of the circular path. There are attempts to clarify the roles of tension and weight in the context of uniform circular motion.

Discussion Status

Some participants have offered insights regarding the nature of forces acting on the mass, suggesting that the force exerted by the rod is not simply tension and may include additional components. There is recognition of a third force that could influence the net force, leading to a deeper exploration of the problem.

Contextual Notes

Participants note the absence of tangential acceleration due to uniform speed, which raises questions about the nature of the net force and its components. The discussion also highlights the complexity of forces involved in the motion, particularly in relation to the assumptions made about tension in a rod versus a string.

twinklestar10
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Homework Statement


An object of mass m is attached to a light rod of length r which is driven by a motor at a steady rate. The object performs uniform vertical circular motion.
Centripetal force is provided by tension and weight.
Draw the net force acting on the mass when the mass is at a height of r from the bottom of the circular path.

Homework Equations


Net force = T + mg


The Attempt at a Solution


so I draw the net force which is pointing inwards, but at the same time, a bit downwards rather than solely towards the centre of the circular path

but isn't the net force of a uniform circular motion pointing towards the centre of the circular path all the time?

Can anyone help me to draw the free body diagram of the mass and explain it to me?
 
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twinklestar10 said:

The Attempt at a Solution


so I draw the net force which is pointing inwards, but at the same time, a bit downwards rather than solely towards the centre of the circular path

Why would it not point direct to the center of the path?
 
twinklestar10 said:
Centripetal force is provided by tension and weight.
.
.
.
Can anyone help me to draw the free body diagram of the mass and explain it to me?
Welcome to PF :smile:

Since it is a rod, not a rope or string, the force it exerts is not simply tension along the rod's length.

p.s. Hello Sylas.
 
but there is no tangential acceleration as the speed is uniform,
so only centripetal acceleration exists...
that means total acceleration is actually centripetal acceleration,
by F=ma, net force should be centripetal too, isn't it?
how come the correct answer isn't like that?
 
twinklestar10 said:
but there is no tangential acceleration as the speed is uniform,
so only centripetal acceleration exists...
that means total acceleration is actually centripetal acceleration,
by F=ma, net force should be centripetal too, isn't it?
Sounds good to me.
how come the correct answer isn't like that?
What do they claim is the correct answer?
 
they claim that...
in the case when the mass is at a height of r
the net force shouldn't be 100% centripetal, but it is at a small angle to the horizontal, say pointing towards 8 o'clock
(please forgive my poor English...)
because there are 2 forces acting on it: weight & tension
I understand this, and from this I can deduce the vector sum of forces which is really not totally centripetal,
but, the speed of the ball is uniform, which means there shouldn't be tangential acceleration, and from this, I can deduce that the total force should only be centripetal...
Did I make any mistakes?
 
You are correct--the book is wrong. Just because two forces act doesn't mean that the net force is not completely centripetal.
 
But it seems not to agree with the free body diagram --- tension + weight,
how can I get the vector sum as a centripetal one?
 
twinklestar10 said:
But it seems not to agree with the free body diagram --- tension + weight,
how can I get the vector sum as a centripetal one?
As Redbelly98 hinted above, the force exerted by the rod is not simply a tension--it can exert a sideways force as well. (Otherwise it couldn't produce uniform vertical circular motion.)
 
  • #10
actually i have an idea...
do you think that the motor would also provide a force directing up (static friction between the rod and the mass?!) in order to move the mass?
that means there are 3 forces,
tension, weight & this upward force
then i can conclude that the vector sum can be 100% centripetal because this upward force balances the weight, what's left is only the tension.
What do you think?
 
  • #11
Doc Al said:
As Redbelly98 hinted above, the force exerted by the rod is not simply a tension--it can exert a sideways force as well. (Otherwise it couldn't produce uniform vertical circular motion.)

Actually, I just saw your reply after I posted my last reply...

Anyway, so I now understand the problem I encountered before,
the key is that I didn't realize there was a third force besides tension, provided by the rod.
and that exactly can cancel the weight when the mass is at a height of r,
this leads to the centripetal net force at that moment.

Thanks a lot to both mentors!
 
  • #12
Hi twinklestar10! :smile:

Forget tension …

a rod doesn't have longitudinal tension.

(If this was a string, the tension would be along the string, but there would be no way to make the motion uniform :wink:)
 
  • #13
[STRIKE]Twinklestar, yes, there is an additional upwards force. It is the sideways force (not tension) that is exerted by the rod on the object. The motor exerts this force on the rod, which in turn exerts the force on the object.[/STRIKE]

EDIT:
Ah, nevermind, you've got it :smile:

twinklestar10 said:
Actually, I just saw your reply after I posted my last reply...

Anyway, so I now understand the problem I encountered before,
the key is that I didn't realize there was a third force besides tension, provided by the rod.
and that exactly can cancel the weight when the mass is at a height of r,
this leads to the centripetal net force at that moment.

Thanks a lot to both mentors!
You're welcome!
 

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