Questions on Rotational Equilibrium

[HopefulOncoDoc]

---

Members don't see this ad.

A solid body can be in rotational equilibrium under all of the following conditions EXCEPT:

A. when it has constant angular momentum.
B. when it experiences no net external torque.
C. when it experiences no net external forces.
D. when it rotates about its center of mass at constant angular velocity.

The answer is C but Torque = r x Fperpendicular so if F = 0, wouldn't Torque = 0 and thus the answer should be either A or D?

A 1 kg meter stick is attached to the ceiling by a string at the 40 cm mark. If a 2 kg mass is attached to the meter stick at the end labeled 0 cm, what mass placed at the 80 cm mark would balance the meter stick?

A. 0.5 kg
B. 1.75 kg
C. 2.0 kg
D. 4.0 kg

The answer is B. Could someone please give a rough calculation on how to go about solving this? Thanks a lot everyone.

 

[RogueUnicorn]

NET force = 0 means the vector sums are zero, not that there is not force. you can have two forces acting in exactly opposite direction and have DOUBLE the torque than a single force alone.

e.g
F1
---> |
.......|
.......|
.......| F2
.......| <----

 

[wanderer]

It still CAN be under rotational equilibrium if there is zero net force. Poor wording there.

 

[ezsanche]

It still CAN be under rotational equilibrium if there is zero net force. Poor wording there.

you are right it can. o well it bet it was a kaplan question.

 

[Omni]

you are right it can. o well it bet it was a kaplan question.

I don't think kaplan is this bad.
However did anyone get question 2? I keep getting 1.8 kg.

 

[Doodl3s]

So, for question two... use the average of the meter stick's weight as separate lengths on each side of the string... So if the whole length is 1 kg, the part of the stick to left of the string weighs .4 kg, and the part to the right weighs .6 kg.

Put the .4 and .6 kg at a distance halfway to the end... so on the left you have (.4 x .2) and on the right you have (.6 x .3)...

Then on the left, add the (2 x .4), and on the right add the (.4 x Y)...

(2 x .4) + (.4 x .2) = (.6 x .3) + (.4 x Y)...

solve for Y = 1.75 😀

 

[wanderer]

I did it like, Tension is unknown so use it as the axis (in other words no lever arm). At .1 m (the center of mass) there is a 1 kg mass, at -.4 there is a 2kg mass, and .4 there is an unknown mass.
.4 * 2 = .4 * X + .1 *1
X = (.8 - .1) / .4 = 1.75 kg

 

[Omni]

So, for question two... use the average of the meter stick's weight as separate lengths on each side of the string... So if the whole length is 1 kg, the part of the stick to left of the string weighs .4 kg, and the part to the right weighs .6 kg.

Put the .4 and .6 kg at a distance halfway to the end... so on the left you have (.4 x .2) and on the right you have (.6 x .3)...

Then on the left, add the (2 x .4), and on the right add the (.4 x Y)...

(2 x .4) + (.4 x .2) = (.6 x .3) + (.4 x Y)...

solve for Y = 1.75 😀

I don't understand where youa re getting the .2 and .3 from. This is how I did it:

Mass1= 2 kg
Partion1= 0.4kg
Distance1= 0.4m

Mass2=y
Partition2=0.6kg
Distance2=0.4m

To solve:
F1 x r1 = F2 x r2
(m1+p1)*g*r1=(m2+p2)*g*r2
(m1+p1)*r1=(m2+p2)*r2

(2+0.4)*0.4=(y+0.6)*0.4

y= [2.4*0.4)/0.4 ]-0.6
y=1.8

 

[HopefulOncoDoc]

I did it like, Tension is unknown so use it as the axis (in other words no lever arm). At .1 m (the center of mass) there is a 1 kg mass, at -.4 there is a 2kg mass, and .4 there is an unknown mass.
.4 * 2 = .4 * X + .1 *1
X = (.8 - .1) / .4 = 1.75 kg

Hey wanderer could you explain how you got the .1?

 

[wanderer]

The center of mass is halfway through the meter stick, or .5 m. (We can imagine the entire mass being concentrated at this single point.) Since I picked the 40 cm mark as the point of 0 lever arm, the 1 kg mass is located .1 m from there.

 

[ezsanche]

 

[RogueUnicorn]

the length people go to help other people on this subforum is quite astounding sometimes.

 

[thebillsfan]

the length people go to help other people on this subforum is quite astounding sometimes.

haha look who's talking...you help people as much as anyone! 👍