To perform an experiment to determine the force constant of a spring, you will need a stand with a boss and clamp, a spiral spring, a meter rule and different weights.
The setup is arranged as shown in the image attached. The natural length of the spring is first recorded. Different weights are added to the spring one after the other and the extension is recorded.
The weight is now plotted on the vertical axis and the extension is plotted on the horizontal axis. The slope of the graph is the force constant of the spring.
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Answer:
T = mg - (m²g/(I/R² + m))
Explanation:
Let T be the tension in the cable between the drum and the bucket
Now, by applying newton's second law of gravity on the downward movement of the bucket, we will obtain;
mg - T = ma - - - - (eq1)
Now, on the drum , a torque of TR will be acting which will create an angular acceleration of "α" in it.
Where R is the radius.
Let "I" denote the moment of inertia of the drum. Thus, we have;
TR = Iα
Now, the angular acceleration is expressed in the form;
α = a/R
Where a is the linear downward acceleration.
Thus;
TR = Ia/ R
T = Ia/ R²
Let's put Ia/ R² for T into equation 1 to give;
mg - Ia/R² = ma
Ia/R² + ma = mg
a( I/R² + m) = mg
a = mg/(I/R² +m)
Now putting mg/(I/R² +m) for a in eq 1 gives;
mg - T = m(mg/(I/R² +m))
T = mg - m(mg/(I/R² +m))
T = mg - m²g/(I/R² + m)