Pendulum Tension

[victorias]

Q. A mass of 1 kg is tied to a 2-meter string and spun in circles at 60 rpm. What is the tension in the string? Assume π = 3.

Answer
The solution gave answer as 72N.
F = ma, and a in this case is v2/r. Since r = 2, we know the circumference is 2*3*2 = 12 meters. At 60 rpm, the mass is traveling 1 revolution per second, or 12 m/s. Hence 122/2 = a; a = 72 m/s2. Since the mass is 1 kg, and F = ma, F = 1 * 72 = 72 N.


Why are they not taking mg into account?

Since no angle is given, I am assuming they are calculating maximum tension at the point when the string is perfectly vertical. Shouldn't we use this equation?

-mg = T + mv^2/r
T = -mg - mv^2/r

---

Members don't see this ad.

 

[aldol16]

Tension is equal to the centripetal force in this case as that is the only force keeping the mass spinning around the circle. Fc = m*v^2/r. Now, since you have m and r, you just have to convert 60 rpm into a linear speed. (60 rotations/minute)*(2pi*r/rotation)*(1min/60 s) = 12 m/s. Therefore, Fc = Ft = (1)(12^2)/(2) = 144/2 = 72 N.

They don't say that they're spinning the string horizontally or vertically. That's a flaw of the question. It could be that they're spinning the string with both horizontal and vertical components in which case you would have to take gravity into account. But they could also be spinning it only horizontally so the centripetal force is the only relevant force acting in this case.

 

[theonlytycrane]

@aldol16 with the given answer, it seems as though it's under the second condition that you described (horizontal spinning only). I get that in this case the centripetal force is the only force keeping the mass spinning around the circle. But what prevents the mass from falling down? I guess I'm picturing this like a tetherball spinning perfectly horizontally around a pole.

 

[victorias]

Okay, so for the vertical pendulum T = mg cos(theta) + mv^2/r

For horizontal pendulum, I am still trying to figure out the forces - can someone please post a free body diagram (visual of what's going on in this case)

 

[aldol16]

@aldol16 with the given answer, it seems as though it's under the second condition that you described (horizontal spinning only). I get that in this case the centripetal force is the only force keeping the mass spinning around the circle. But what prevents the mass from falling down? I guess I'm picturing this like a tetherball spinning perfectly horizontally around a pole.

It couldn't be perfectly horizontal and spinning the air because as you have figured out, that would mean forces in the vertical direction are not balanced. So there are two explanations. The first and less likely one is that you can imagine it spinning on a frictionless surface so that the normal force balances the force of gravity in the vertical direction. This will work for this problem. Or, you can imagine that the angle the string makes with the horizontal is very very small, i.e. approaching zero. That means that in the horizontal direction, you have the component of tension which provides centripetal force. This would be tension force multiplied by the cosine of the angle the string made with the horizontal. Remember that we assumed this was very small. Therefore, Ft*cos(theta) = Ft when theta is very small because that means that cos(theta) would approach unity.

Again, I see this as a fundamental flaw in the question because it should either tell you that it's spinning horizontally or show that to you in a figure. Without that information, you can't safely make that assumption so it's not a good MCAT question.

 

[victorias]

Thanks!

Another question about pendulums. In the figure below, I am confused with the x and y label they have put on the components of Fg - they seem switched around to me??

 

[aldol16]

It doesn't matter how you define your coordinate system as long as you're consistent. In other words, x and y are just names that we use to call variables. It works just as well if you switch the names.