bucket and pulley

[logistic_guy]

A uniform solid cylindrical pulley of mass \(\displaystyle M = 3 \ \text{kg}\) and radius \(\displaystyle 20 \ \text{cm}\) is mounted on a horizontal axle. The pulley is free to rotate, but a constant frictional torque of \(\displaystyle \tau_f = 0.6 \ \text{Nm}\) acts at the axle and opposes the pulley’s rotation. A light, inextensible, and non-stretching rope is tightly wrapped around the pulley. The rope has negligible mass and does not slip as it unwinds. A bucket of mass \(\displaystyle m = 4 \ \text{kg}\) is attached to the free end of the rope and hangs vertically. The system is released from rest at time \(\displaystyle t = 0\) and the bucket begins to fall.

Determine:

\(\displaystyle \bold{(a)}\) The moment of inertia \(\displaystyle I\) of the pulley.
\(\displaystyle \bold{(b)}\) The linear acceleration \(\displaystyle a\) of the bucket.
\(\displaystyle \bold{(c)}\) The tension \(\displaystyle T\) in the rope.
\(\displaystyle \bold{(d)}\) The angular acceleration \(\displaystyle \alpha\) of the pulley.
\(\displaystyle \bold{(e)}\) The angular velocity \(\displaystyle \omega\) of the pulley at \(\displaystyle t = 2 \ \text{s}\).
\(\displaystyle \bold{(f)}\) The linear velocity \(\displaystyle v\) of the bucket at \(\displaystyle t = 2 \ \text{s}\).

 

[khansaheb]

A uniform solid cylindrical pulley of mass \(\displaystyle M = 3 \ \text{kg}\) and radius \(\displaystyle 20 \ \text{cm}\) is mounted on a horizontal axle. The pulley is free to rotate, but a constant frictional torque of \(\displaystyle \tau_f = 0.6 \ \text{Nm}\) acts at the axle and opposes the pulley’s rotation. A light, inextensible, and non-stretching rope is tightly wrapped around the pulley. The rope has negligible mass and does not slip as it unwinds. A bucket of mass \(\displaystyle m = 4 \ \text{kg}\) is attached to the free end of the rope and hangs vertically. The system is released from rest at time \(\displaystyle t = 0\) and the bucket begins to fall.

Determine:

\(\displaystyle \bold{(a)}\) The moment of inertia \(\displaystyle I\) of the pulley.
\(\displaystyle \bold{(b)}\) The linear acceleration \(\displaystyle a\) of the bucket.
\(\displaystyle \bold{(c)}\) The tension \(\displaystyle T\) in the rope.
\(\displaystyle \bold{(d)}\) The angular acceleration \(\displaystyle \alpha\) of the pulley.
\(\displaystyle \bold{(e)}\) The angular velocity \(\displaystyle \omega\) of the pulley at \(\displaystyle t = 2 \ \text{s}\).
\(\displaystyle \bold{(f)}\) The linear velocity \(\displaystyle v\) of the bucket at \(\displaystyle t = 2 \ \text{s}\).

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[logistic_guy]

\(\displaystyle \bold{(a)}\) The moment of inertia \(\displaystyle I\) of the pulley.

We are now experts in moments of inertia and we have already derived the moment of inertia of a cylinder about its central axis.

The moment of inertia of the pulley is:

\(\displaystyle I = \frac{1}{2}MR^2 = \frac{1}{2}(3)(0.2)^2 = \textcolor{blue}{0.06 \ \text{kg} \cdot \text{m}^2}\)

 

[logistic_guy]

\(\displaystyle \bold{(b)}\) The linear acceleration \(\displaystyle a\) of the bucket.

Gravity is pulling the bucket downward, so we can get our first equation as:

\(\displaystyle mg - T = ma\)

Two unknowns means that we cannot find the linear acceleration of the bucket right now. We then go to the other system. From the pulley, we can get:

\(\displaystyle RT - \tau_f = I\alpha = I\frac{a}{R}\)

Now we can take the tension in the rope \(\displaystyle T\) from the first equation and substitute it in the second equation to get the linear acceleration \(\displaystyle a\).

\(\displaystyle T = mg - ma\)

\(\displaystyle R(mg - ma) - \tau_f = I\frac{a}{R}\)

\(\displaystyle R^2mg - R^2ma - R\tau_f = Ia\)

\(\displaystyle Ia + R^2ma = R^2mg - R\tau_f \)

\(\displaystyle a = \frac{R^2mg - R\tau_f}{I + R^2m} = \frac{(0.2)^2(4)(9.8) - (0.2)(0.6)}{0.06 + (0.2)^2(4)} = \textcolor{blue}{6.58 \ \text{m}/\text{s}^2}\)

 

[logistic_guy]

\(\displaystyle \bold{(c)}\) The tension \(\displaystyle T\) in the rope.

\(\displaystyle T = mg - ma = m(g - a) = 4(9.8 - 6.58) = \textcolor{blue}{12.88 \ \text{N}}\)

 

[logistic_guy]

\(\displaystyle \bold{(d)}\) The angular acceleration \(\displaystyle \alpha\) of the pulley.

\(\displaystyle a = \alpha R\)

\(\displaystyle \alpha = \frac{a}{R} = \frac{6.58}{0.2} = \textcolor{blue}{32.9 \ \text{rad/}\text{s}^2}\)

 

[logistic_guy]

\(\displaystyle \bold{(e)}\) The angular velocity \(\displaystyle \omega\) of the pulley at \(\displaystyle t = 2 \ \text{s}\).

\(\displaystyle \omega = \omega_0 + \alpha t\)

\(\displaystyle \omega = 0 + (32.9)(2) = \textcolor{blue}{65.8 \ \text{rad/s}}\)

 

[logistic_guy]

\(\displaystyle \bold{(f)}\) The linear velocity \(\displaystyle v\) of the bucket at \(\displaystyle t = 2 \ \text{s}\).

\(\displaystyle v = v_0 + at\)

\(\displaystyle v = 0 + (6.58)(2) = \textcolor{blue}{13.16 \ \text{m/s}}\)