TBR physics Ch 2 passage II Question 14

[silverice]

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The experiment (wording of the question is slightly modified from the TBR question, but the meaning of the question remain the same)
The student takes the ball and string and swings the ball in a vertical circle such that the ball nearly hits the floor. The length of the string is L.

Question 13
when the ball is directly above the ground during the swing, the tension in the string is measured to be three times the weight of the ball. what is the speed of the ball?
A. 2sqrt root (gL)
B. sqrt root (3gL)
C. sqrt root (2gL)
D. 2gL


I set Tension equal to mg, T=3mg
then ma=3mg
then mv^2/r =3mg
and got choice B

I guess i'm having trouble with the concept of centripetal force. I thought the tension of the string is centripetal force because it is pulling the object toward the center.

Thank you for your helps

 

[ljc]

So what is the correct answer? And the book's explanation?

 

[silverice]

So what is the correct answer? And the book's explanation?

the book's answer is A

Fcentripetal = T + mg= (mv^2)/L T=3mg

3mg + mg =(mv^2)/L solve for V

The tension in the string is measured to be three times the weight of the ball. To calculate the speed of the ball at this point, we must use Newton's second law.

 

[Ssina]

F centripetal is a NET force, not the tension, so you have to consider mg plus the tension as Fnet

 

[MedPR]

F centripetal is a NET force, not the tension, so you have to consider mg plus the tension as Fnet

This. Since force is equal and opposite, it includes the tension and the mg.

 

[silverice]

This. Since force is equal and opposite, it includes the tension and the mg.

I thought the tension is the Fc, because it's holding the object, and the direction is toward teh center, isn't it?

 

[typicalindian]

I did

net forces = ma
forces are : + T, -W so you get: T - W = ma

3w - w = ma
3mg - mg = m * v^2/L
now solve for v and I got sqrt(2gL) option C...am I mathematically challenged or something? :x

 

[MT Headed]

I like typicalindian's response. At the bottom of the swing the tension is up, and gravity is (always) down. One mg cancels out the gravity, that only leaves two more mg's to do circular motion at that point in the swing.

Fnet = ma
2mg = m(v^2/L)

Answer C is looking pretty tasty to me too.

hmm... when they say "the ball is directly above the ground" do they mean the top of the swing or the bottom of the swing? Because if they meant the top of the swing, then yeah answer A is correct. Seems kind of ambiguous to me.

 

[MrNeuro]

I did

net forces = ma
forces are : + T, -W so you get: T - W = ma

3w - w = ma
3mg - mg = m * v^2/L
now solve for v and I got sqrt(2gL) option C...am I mathematically challenged or something? :x

The question clearly states in TBR that its at point A and point A is the TOP so it would be

3w + w = ma
3mg + mg = m v^2/r
sqrt 4gr = v
v= 2 sqrt gr

r= radius of circle = L hence answer A

I like typicalindian's response. At the bottom of the swing the tension is up, and gravity is (always) down. One mg cancels out the gravity, that only leaves two more mg's to do circular motion at that point in the swing.

Fnet = ma
2mg = m(v^2/L)

Answer C is looking pretty tasty to me too.

hmm... when they say "the ball is directly above the ground" do they mean the top of the swing or the bottom of the swing? Because if they meant the top of the swing, then yeah answer A is correct. Seems kind of ambiguous to me.

👍 the question says

At point A in Experiment 2, the tension in the string is measured to be three times the weight of the ball. What is the speed of the ball

point A is the "maximum height" of the circle so the centripetal force is as you said Tension plus weight = mv^2/r

 

[typicalindian]

The question clearly states in TBR that its at point A and point A is the TOP so it would be

3w + w = ma
3mg + mg = m v^2/r
sqrt 4gr = v
v= 2 sqrt gr

r= radius of circle = L hence answer A



👍 the question says

At point A in Experiment 2, the tension in the string is measured to be three times the weight of the ball. What is the speed of the ball

point A is the "maximum height" of the circle so the centripetal force is as you said Tension plus weight = mv^2/r

oh ok I hadn't even looked at the question so I was just going off of how OP phrased the question, either way I had the concept down

 

[qw098]

I like typicalindian's response. At the bottom of the swing the tension is up.

Is tension always up at the bottom of the swing? How do I know in what direction tension is when I am swinging something around?

I know BR says at the top of the swing tension is pointing down. But why is that?

I have attached a picture:

Because tension is going towards the center of the string that is being spun around, is that the same reason why in the picture below.. the normal force is towards the center? Why is Normal force towards the center in the picture below?