Must There Be Net Zero Force and Torque on a Massless Rod?

[timetraveller123]

Homework Statement

i know problems of this kind have been posted a lot but i have a few of my own questions please take a look at them thanks

i am having several conceptual errors in rotational dynamics
one of them is :
since the rod is massless must there be net zero force and torque on it?

Homework Equations

T = I α
f= ma

The Attempt at a Solution

a) is easy
$x^2 + y^2 = l^2\\ {\dot x}^2 + x \ddot x = -{\dot y}^2 - y\ddot y$
the problem comes in at b
i know the centre of mass traces out a circle of radius L/2 about the origin

i do not how to proceed
--like from many websites states N= 2mg but would that be so as the centre of mass is surely coming down then why must net vertical force be zero?
--please help i am not being lazy i truly need help

 

[haruspex]

a) is easy

I would think the question wants an answer involving only $\theta, \ddot x, \ddot y, l$.

many websites states N= 2mg but would that be so as the centre of mass is surely coming down

I agree with you, it is not 2mg.
Let the force in the rod be F. What equations can you write for the two accelerations?

 

[haruspex]

I would think the question wants an answer involving only $\theta, \ddot x, \ddot y, l$.

On second thoughts, I don't think that is possible. E.g. consider y at constant speed. If the speed is zero then x is not accelerating, but if the speed is nonzero then x may be accelerating.
Maybe involve $\dot\theta$ instead of $\dot x$ and $\dot y$. That gets it down to one speed variable while maintaining symmetry between x and y.

 

[timetraveller123]

ow could i do that this is what i get

$y = x tan \theta\\ \dot y = \dot x tan\theta + x \dot \theta sec^2 \theta \\ \ddot y = \ddot x tan \theta + 2 \dot x \dot \theta sec^2 \theta + 2 x \dot \theta sec^2 \theta tan \theta + x \ddot \theta sec^2 \theta$
this is what i got how to eliminate x variables

 

[timetraveller123]

about the problem itself

$m a_{cm,y} = N- 2mg\\ m a_ {cm,x} = R\\$
energy equation no rotational terms as massless rods
$mgl = mgl sin \theta + \frac{1}{2}m (\dot y)^2 + \frac{1}{2} m(\dot x )^2\\ 2gl(1 - sin \theta) = (\dot y)^2+ (\dot x)^2$
i am really confused about where am i going some direction may help
what is the strategy for this question

 

[haruspex]

ow could i do that

Start with y=Lsin(θ) etc.

 

[haruspex]

what is the strategy for this question

Did you try what I suggested in post #2?

 

[timetraveller123]

Start with y=Lsin(θ) etc.

tell if this is right or wrong

$y = L sin \theta \\ \dot y = L\dot \theta cos \theta\\ \ddot y = - L {\dot \theta}^2 sin \theta + L\ddot \theta cos \theta \\ x = L cos \theta \\ \dot x = -L \dot \theta sin \theta \\ \ddot x = - L {\dot \theta}^2 cos \theta - L \ddot \theta sin \theta$
is this correct thanks

 

[timetraveller123]

Did you try what I suggested in post #2?

i already did in post #5
1 equation for vertical motion of com only has normal and gravity acting vertically
another equation for horizontal motion for com only has normal force R from the wall
is that the two equations?

 

[haruspex]

i already did in post #5
1 equation for vertical motion of com only has normal and gravity acting vertically
another equation for horizontal motion for com only has normal force R from the wall
is that the two equations?

No. The Y mass does not directly feel N, it feels the vertical component of the force in the rod. Similarly the X mass feels the horizontal component of it.

 

[timetraveller123]

ok now i see
is it
$Fsin\theta - mg= \ddot y\\ Fcos \theta = \ddot x$

 

[haruspex]

ok now i see
is it
$Fsin\theta - mg= \ddot y\\ Fcos \theta = \ddot x$

Right, except you forgot the m on the right in each.

 

[timetraveller123]

i am getting

$\ddot xtan \theta - g = \ddot y$
initially theta is 90 but tan 90 blows up

 

[haruspex]

i am getting

$\ddot xtan \theta - g = \ddot y$
initially theta is 90 but tan 90 blows up

If the initial position were upright there wouldbe no horizontal force on the lower mass, so no movement.
The initial condition is with some θ=θ₀, but $\dot \theta=0$.

 

[timetraveller123]

give me some time i am currently studying a chapter in a book about angular momentum and rigid body dynamics i think after studying that i should better outlook on this problem i will get back to you soon i also have follow up problem to this thanks for your understanding

 

[timetraveller123]

sorry for the late reply
so as you said
inital acceleration
$\ddot xtan \theta_o - g = \ddot y$
(doubt)would the horizontal acceleration be the same for both masses?

 

[haruspex]

would the horizontal acceleration be the same for both m

No, one moves only vertically, the other only horizontally.
If the descending mass is free to leave the wall then that would happen eventually, but you are only interested in the initial movement.