Understanding Centripetal Force in Circular Motion Experiments

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In the circular motion experiment, the centripetal force is provided by the tension in the string, which is equal to the weight of the mass hanging from it. The formula for centripetal force is f = mv²/r, but in this scenario, the tension (T) in the string serves as the centripetal force. Since the setup involves a frictionless loop, the tension remains constant throughout the string. Therefore, the centripetal force can be calculated as T = mg, where m is the mass and g is the acceleration due to gravity. This understanding clarifies the relationship between mass, tension, and centripetal force in the experiment.
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Homework Statement


I did this circular motion experiment where we had to change the mass of weights on the bottom of a fishing wire and we kept the radius the same. The questions in the post lab say to graph "centripetal force vs velocity". I worked out the velocity by using the 2pi*r/t formula but I'm not sure about the centripetal force. Would it be the mass on the bottom of the string providing the centripetal force? Very confused and would appreciate some help :confused:

Thanks


Homework Equations





The Attempt at a Solution


 
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The formula for clculating centripetal force is:

f =\frac{mv^{2}}{r}

r is the radius from the axis of rotation to the mass.
 
rollcast said:
The formula for clculating centripetal force is:

f =\frac{mv^{2}}{r}

r is the radius from the axis of rotation to the mass.

I know this is the formula but apparently in this experiment, I'm supposed be only using the mass to find the centripetal force because the masses are hanging off the bottom of the string while the top is being whirled around.
 
hi paperdoll! :smile:
paperdoll said:
I know this is the formula but apparently in this experiment, I'm supposed be only using the mass to find the centripetal force because the masses are hanging off the bottom of the string while the top is being whirled around.

oh, you mean the string goes from a mass up to the centre, through a frictionless loop, then carries on to a weight that's being whirled around?

so the mass stays still?

in that case, the centripetal force is the tension T in the string

since the tension is the same throughout the string (because the loop is frictionless), it's also the only force that's holding up the mass …

so doing F = ma on the mass we have T - mg = 0,

so in this case, yes, the centripetal force (= T) is mg :wink:
 
tiny-tim said:
hi paperdoll! :smile:


oh, you mean the string goes from a mass up to the centre, through a frictionless loop, then carries on to a weight that's being whirled around?

so the mass stays still?

in that case, the centripetal force is the tension T in the string

since the tension is the same throughout the string (because the loop is frictionless), it's also the only force that's holding up the mass …

so doing F = ma on the mass we have T - mg = 0,

so in this case, yes, the centripetal force (= T) is mg :wink:


Thanks! That clears things up now :)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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