A 230g mass hangs from a string that is wrapped around a pulley. The pulley is suspended in such a way that it can rotate freely. When the mass is released, it accelerates toward the floor as the...

Question

A 230g mass hangs from a string that is wrapped around a pulley. The pulley is suspended in such a way that it can rotate freely. When the mass is released, it accelerates toward the floor as the string unwinds. Model the pulley as a uniform solid cylinder of mass 1.00 kg and radius 7.00 cm. Assume the thread has negligible mass and does not slip or stretch as it unwinds.

Determine the magnitude 𝛼 of the pulley's angular acceleration.

Determine the magnitude of the acceleration 𝑎 of the descending weight.

Calculate the magnitude of the tension 𝑇 in the string.

Answer 1

Let

m = 230 g = 0.23 kg be the mass of the weight,

M = 1 kg be the mass of the pulley,

r = 7 cm = 0.07 m be the radius of the pulley,

I be the moment of inertia of the pulley,

T (to be calculated) be the tension on the string,

a (to be calculated) be the acceleration of the weight,

α (to be calculated) be the angular acceleration of the pulley,

g = 9.81 m/s² be gravitational acceleration.

The pulley experiences only one force -- the downward force of tension T.

As the tension acts at radius r, it creates torque of rT.

By Newton's second law applied to rotation:

rT = αI (1)

The weight experiences two forces --

the downward force mg of gravity, and the upward force of tension T.

By Newton's second law

mg - T = ma (2)

There is a relationship between angular acceleration α and linear acceleration a:

a = αr (3)

A uniform disk has the moment of inertia

I = Mr²/2 (4)

This gives us four equations with four unknowns (T, a, α, I) to solve.

From (2)

T = mg - ma (5)

Substitute (3), (4), (5) into (1)

r(mg - mrα) = αMr²/2

Express

α = mg/r(m + M/2)

From(3)

a = mg/(m + M/2)

From (5)

T = mg(1 - m/(m + M/2))

The result can be interpreted as follows.

In general, acceleration is the ration between a force and resistance to acceleration.

In our case, the only force causing movement is gravity -- mg.

Resistance to acceleration is mass -- m and M.

The pulley contributes only half of its mass M to the resistance because the mass

is not all at the perimeter -- it is distributed throughout the pully's body.

Substituting actual numbers

a = 0.23×9.81/(0.23 + 1/2) = 3.09 m/s²

α = a/r = 3.09/0.07 = 44.15 s^-2

T = mg - ma = m(g - a) = 0.23×(9.81 - 3.09) = 1.55 N