A bucket of mass m is hanging from the free end of a rope whose other end is wrapped around a drum (radius R, mass M) that can r
1 answer:
Response:
Explanation:
Let T denote the tension.
By employing Newton's second law to analyze the bucket's downward motion, we have:
mg - T = ma
A torque, TR, acts on the drum, inducing an angular acceleration α in it. If I refers to the moment of inertia of the drum, then:
TR = Iα
Rearranging gives: TR = Ia/R
This leads to T = Ia/R²
Substituting this expression for T back into the previous equation yields:
mg - T = ma
mg - Ia/R² = ma
Consequently, we find that mg = Ia/R² + ma
Therefore, a (I/R² + m) = mg
This results in: a = mg / (I/R² + m)
Next, we aim to express T as:
mg - T = ma
which simplifies to mg - ma = T
Rearranging gives mg - m²g / (I/R² + m) = T
Thus, we arrive at: mg - mg / (1 + I / m R²) = T
For part (b), T = Ia/R²
and for part (c), the moment of inertia of a hollow cylinder calculates to:
I = 1/2 M (R² - (R² / 4))
This simplifies to 3/4 x 1/2 MR², yielding 3/8 MR²
Thus, I / R² = 3/8 M
When we substitute, we find a = mg / (3/8 M + m)
and subsequently T = Ia/R²
= 3/8 MR² × mg / (3/8 M + m) × 1/R²
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Answer:
The period of the pendulum measuring 16 m is double that of the 4 m pendulum.
Explanation:
Recall that the period (T) of a pendulum with length (L) is defined by:
where "g" denotes the local gravitational acceleration.
Since both pendulums are positioned at the same location, the value of "g" will be consistent for both, and when we compare the periods, we find:
Thus, the duration of the 16 m pendulum is two times that of the 4 m one.