Question

A light flexible rope is wrapped several times around a solid cylinder of mass 50 kg and diameter 0.12m, which rotates without friction about fixed horizontal axis. The free the end of the rope is pulled with a constant force of magnitude 9N for a distance of 2m. If the cylinder was initially at rest, find its final angular velocity and the final speed of the rope.

Answer 1

A light flexible rope is wrapped several times around a solid cylinder of mass 50 kg and diameter 0.12 m, which rotates without friction about fixed horizontal axis. The free the end of the rope is pulled with a constant force of magnitude 9N for a distance of 2 m. If the cylinder was initially at rest, find its final angular velocity and the final speed of the rope.

Newton's Laws for Rotation:

$$I \beta = F * \left(\frac {d}{2}\right)$$

I – moment of inertia, $I = m \left( \frac{d}{2} \right)^2 = \frac{m d^2}{4}$

$\beta$ – angular acceleration

$$\beta = F * \frac {\left(\frac {d}{2}\right)}{m \left(\frac {d}{2}\right) ^ {2}} = \frac {2 F}{m d}$$

Angular velocity equals:

$$\omega = \beta t \quad \Rightarrow \quad t = \frac {\omega}{\beta}$$

And angle equals:

$$\varphi = \frac {\beta t ^ {2}}{2} = \frac {\omega^ {2}}{2 \beta}$$

Distance equals:

$$l = \varphi * \left(\frac {d}{2}\right) \quad \Rightarrow \quad \varphi = \frac {2 l}{d}$$

Therefore:

$$\begin{array}{l} 2 \frac {l}{d} = \frac {\omega^ {2}}{2 \beta} \quad \Rightarrow \quad \omega = 2 \sqrt {\frac {\beta l}{d}} = 2 \sqrt {\frac {2 F}{m d} \frac {l}{d}} = 2 \sqrt {\frac {2 F l}{m d ^ {2}}} \\ \omega = 2 \sqrt {\frac {2 F l}{m d ^ {2}}} = 14.14 \frac {rad}{sec} \\ \end{array}$$

Final speed of rope equals: $v = \omega * \left(\frac{d}{2}\right) = \sqrt{\frac{2Fl}{m}} = \frac{3\sqrt{2}}{5} = 0.85\frac{m}{s}$

Answer: $\omega = 14.14\frac{rad}{sec}, v = 0.85\frac{m}{s}$