The solution of a problem regarding center of mass

[Likith D]

A rod of mass M (=4m) and length L is placed over a smooth horizontal table. One end of rod is attached with a block of mass m through a light string as shown in the figure. The pulley is friction less and light. the acceleration of com of rod at the given instant is (g = acceleration due to gravity)

a) g/2
b) g/8
c) g/5
d) 2g/5

that does not make sense, because i think the force on com is net force on object = mg = Ma (force of com by Newton's law)
(M = mass of com = 4m)
therefore, a = mg/M = mg/4m = g/4 which is none of the above

note:
1. Ma = mg is true even if the body rotates
2. It's reasonable to have no friction, since it is smooth
3. The rod does not seem to have any of it's points fixed to table to revolve around of
4. There was no use of the fact that force was perpendicular to rod as in the figure

 

[Orodruin]

What makes you think the force is equal to mg?

 

[Likith D]

What makes you think the force is equal to mg?

yeah, you are right, that must have been tension... but how do you find the tension in the rope
If i were to hold the string wouldn't i have to exert mg to hold it in place... thus the force exerted on me which is it's tension is mg... why would that be different in this case where there is a rod instead

 

[Orodruin]

You need to somehow relate the motion of the rod (both linear and rotational acceleration) to that of the hanging mass.

The easy part is: how does the tension in the rope relate to the acceleration of the hanging mass? See if you can figure out what other questions are helpfulto ask.

 

[Likith D]

You need to somehow relate the motion of the rod (both linear and rotational acceleration) to that of the hanging mass.

The easy part is: how does the tension in the rope relate to the acceleration of the hanging mass? See if you can figure out what other questions are helpfulto ask.

acceleration of hanging body is a = (g - T/m)
that's what i make of one part of tension...
concerning the rod...
A(com) = T/4m

(about com) = (L/2)*(T) = Iα
therefore α = (LT/2)/(ML^2/12) = 3T/2mL

woa... does the solution have anything to do with the fact that length of string does not change?

 

[Orodruin]

So how can you use these relations to find the tension (and hence the CoM acceleration)?

Hint: You need one more relation that relates the CoM acceleration, the angular acceleration, and the acceleration of the hanging mass as you currently have 3 equations (a = g - T/m, A = T/4m, and α = 3T/2mL) and 4 unknowns (a, T, A, and α).

 

[Likith D]

So how can you use these relations to find the tension (and hence the CoM acceleration)?

Hint: You need one more relation that relates the CoM acceleration, the angular acceleration, and the acceleration of the hanging mass as you currently have 3 equations (a = g - T/m, A = T/4m, and α = 3T/2mL) and 4 unknowns (a, T, A, and α).

Does it require the use of energy conservation?
or does it come from the fact that the string length is constant?
It doesn't seem to come from energy because it will introduce height, a new variable, right?

 

[Orodruin]

What do you get if you use that the string length is constant?

 

[Likith D]

What do you get if you use that the string length is constant?

but that makes acceleration of hanging mass and tip of the rod the same? so that they make the rope constant in length...
then i get a = Lα /2
Is that the fourth equation i needed?

 

[Orodruin]

but that makes acceleration of hanging mass and tip of the rod the same?

No, the acceleration of the rod CoM does not need to equal the acceleration of the hanging mass as the rod can turn. The acceleration of the tip of the rod however ...