Centripetal motion, the tension at the bottom of a circle

In summary: Yes. Note that 6mg and 5mg were the only options that could possibly result in the same total energy, since they both have 1/2(mLg) = (m)v^2 in them.
  • #1
Kennedy
70
2

Homework Statement


A mass m, at one end of a string of length L, rotates in a vertical circle just fast enough to prevent the string from going slack at the top of the circle. Assuming mechanical energy is conserved, the tension in the string at the bottom of the circle is: a) 6 mg b) mg + 3mg/L c) 5 mg d) 5gL e) 4 mgL

Homework Equations


a = v^2/r
KEi + PEi = KEf + KEi

The Attempt at a Solution


I understand that the centripetal acceleration at the top of the circle is 9.8 m/s^2, because then there is no tension in the string because the centripetal force all comes from the weight of the object. I understand that the PE at the top of the circle is 2L(g)(m), considering that the bottom of the circle is 0 PE. Now, since energy is conserved, the speed of the mass at the bottom of the circle is found by using the fact that the negative of the change in potential energy is equal to the kinetic energy. At the bottom of the circle, all of the potential energy is converted into kinetic energy. so, 2(L)(g)(m) = 1/2(m)(v^2). This yields v = (4Lg)^(1/2). So, to find the centripetal force at the bottom of the motion (otherwise known as the tension in the string) a = 4Lg/L = 4g, then Fc = m(4g) = 4mg, but clearly this isn't an option, but the answer that makes the most sense to me.
 
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  • #2
Is the mass at rest at the top of the circle?

Is the tension equal to the centripetal force at the bottom?

Edit: You should also be able to discard several alternatives without doing any computations solely based on dimensional arguments.
 
  • #3
Orodruin said:
Is the mass at rest at the top of the circle?

Is the tension equal to the centripetal force at the bottom?
Well, no. Because then the centripetal acceleration would be equal to zero (ac = v^2/r), and if the velocity is zero then there is no centripetal acceleration, and hence no circular motion.
 
  • #4
So why are you assuming that there is no kinetic energy at the top? Also, you did not answer the second question.
 
  • #5
Orodruin said:
So why are you assuming that there is no kinetic energy at the top? Also, you did not answer the second question.
Ooohhh... Okay. So the total energy is not only due to potential energy. The speed of the mass at the top would be (gr)^(1/2), which yields a total energy of 2Lmg + 1/2(m)(g)(L) = (5/2)mLg. Which would then all be converted into KE at the bottom, so (5/2)mLg = 1/2(m)v^2. The speed at the bottom would be (5Lg)^(1/2). Then this means that the centripetal acceleration would be 5Lg/L = 5g, and the centripetal force would be 5mg, but according to my answer key this is still not the right answer.

...and to answer your second question, yes. I think the centripetal force is equal to the tension of the string at the bottom, but since you ask, I assume that's wrong.
 
  • #6
Kennedy said:
...and to answer your second question, yes. I think the centripetal force is equal to the tension of the string at the bottom, but since you ask, I assume that's wrong
Indeed. Now, why could it be wrong? Have you drawn the free body diagram for the mass at the bottom? What forces act on it? Is the tension equal to the centripetal force at the top? Why/why not?
 
  • #7
Orodruin said:
Indeed. Now, why could it be wrong? Have you drawn the free body diagram for the mass at the bottom? What forces act on it? Is the tension equal to the centripetal force at the top? Why/why not?
Is it because the weight of the object at the bottom is exerting a force downwards, and to "compensate" for the force acting downwards, the tension in the string needs to be the centripetal force plus the weight of the object? This would give 5mg + mg = 6mg? Is that the right logic behind the problem?

Now, to clarify, because energy is conserved here, the centripetal force changes constantly throughout the circular motion? Because normally with circular motion problems the tension in the string at the bottom is simply the centripetal force of the string (which is always the same) plus the weight of the object. Except in this case, the centripetal force at the top of the circle is mg, whereas at the bottom it is 5mg. Why does this happen?
 
  • #8
Kennedy said:
Is it because the weight of the object at the bottom is exerting a force downwards, and to "compensate" for the force acting downwards, the tension in the string needs to be the centripetal force plus the weight of the object? This would give 5mg + mg = 6mg? Is that the right logic behind the problem?
Yes. Note that 6mg and 5mg were the only options that could possibly describe a tension. All other options had the wrong physical dimensions.

Kennedy said:
Because normally with circular motion
I would not say ”normally” here. It is true only if you have a constant velocity, which you do not here.

Kennedy said:
Except in this case, the centripetal force at the top of the circle is mg, whereas at the bottom it is 5mg. Why does this happen?
Not ”except” in this case, in any case where the speed is not constant, which is most cases. Why does it happen here? You have solved the problem so you should be able to tell me why the speed increased.
 

Related to Centripetal motion, the tension at the bottom of a circle

What is centripetal motion?

Centripetal motion is the motion of an object along a circular path at a constant speed. It is caused by a centripetal force, which is directed towards the center of the circle.

What is the tension at the bottom of a circle?

The tension at the bottom of a circle is the force exerted by a string, rope, or other object that is keeping an object in circular motion. It is equal to the centripetal force required to maintain the circular motion.

How is the tension at the bottom of a circle calculated?

The tension at the bottom of a circle can be calculated using the equation T = mv²/r, where T is the tension, m is the mass of the object, v is the velocity, and r is the radius of the circle.

What happens to the tension at the bottom of a circle if the speed increases?

If the speed increases, the tension at the bottom of the circle will also increase. This is because the centripetal force required to maintain the circular motion increases with the speed.

Can the tension at the bottom of a circle be greater than the weight of the object?

Yes, the tension at the bottom of a circle can be greater than the weight of the object. This can happen if the object is moving at a high enough speed or if the radius of the circle is small enough. In this case, the centripetal force will be greater than the force of gravity and the object will continue to move in a circular path.

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