TBR Physics Passage Question

[nycneighbor]

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For problem 7 in TBR section 5, it asks:

What is the tension in the string in Trial 4 if the pendulum bob swings through equilibrium with a speed of 4 m/s? (The only things from Trial 4 is that the length is 4 m, and the period is 4 seconds, neither of which matter to the problem.)

A) 0.4 N
B) 0.6 N
C) 1.0 N
D) 1.4 N

The solution says D, but I'm wary of this.

It says that in a free body diagram of the pendulum at equilibrium, tension points upwards, and weight points downwards. It then says:

Sum Fy = T - mg = mv^2/r

And then solves for T = 1.4 N.

Aren't mv^2/r and T pointing in the same direction? And just conceptually, how can the tension of a string be larger than the weight if the pendulum is at equilibrium?

 

[VanBrown]

no, angluar velocity points tangent to the path. Tension always points away from a mass along the rope or whatever is connecting it. However, in this problem you have more than just weight acting in the vertical direction, you also have centripetal force. Therefore the tension will equal the weight (since at equilibrium the W = T) and the additional force created by centripetal acceleration. This force points in the direction of T but don't let that confuse you as to how it affects this problem. Centripetal force is real but its counterpart centrifugal force is NOT but it is the equal and opposite force required by Newtons 3rd (?) law which adds to the apparent weight. Think of being on a tilt-o-whirl. You are being pushed outwards and your apparent weight is increased.

so at the point of equilibrium you have

T=mg + (mv^2)/r

I don't have the book you took the question from but it should have given you mass of the pendulum...

 

[Abe468]

Edit: I just realized I posted this to a really old thread...oh well, anyone see a flaw in my reasoning?

Ya, it gives the mass of the ball as 100g in the passage. I was a little confused by the way they drew the free body diagram as well since I thought centripetal always pointed inward, but they were showing (mv^2)/r going away from the center of the circle. VanBrown's response kind of makes sense to me.

The way I thought of it though was that the net force is equal to mass * acceleration, which for this system is (mv^2)/r. The net force is also equal to the sum of the individual forces. Taking up to be the positive direction that says that:
netF = Tension - mg
and then since the netF = (mv^2)/r
(mv^2)/r = T - mg
This might not be the best way to think about it though. Just what I did when I tried making sense of it for myself.