Question

a pulley hangs from a string, one end of which is held in place on the ceiling, while the other end winds around a spool. the spool and the pulley are made from identical uniform solid disks. the pulley is released from rest, unwinding the spool as it descends, and naturally the string never slips over the surface of the pulley. find the linear speed in meters per second of the pulley at the moment it has fallen a distance of 2.25m. hint: compare the speed of the string coming off the spool to the speed of the falling pulley. provide at least two decimal places

Explanation:

Step1: Analyze the motion relationship

Let the linear speed of the pulley be \(v\). For a solid - disk pulley (rolling without slipping), if the radius of the pulley is \(r\), the angular speed of the pulley \(\omega=\frac{v}{r}\). Since the string unwinds from the spool (which is also a solid disk), and considering the motion of the string, the speed of the string coming off the spool is related to the angular speed of the spool. If the radius of the spool is the same as that of the pulley (because they are identical disks), and the spool rotates as the string unwinds, we know that the speed of the string \(v_s\) and the linear speed of the pulley \(v\) are related. The speed of the string coming off the spool is \(v_s=\frac{v}{2}\) (by considering the kinematics of the double - pulley system). The angular speed of the spool \(\omega_s=\frac{v_s}{r}=\frac{v}{2r}\)

Step2: Apply the conservation of mechanical energy

The initial mechanical energy \(E_i = 0\) (released from rest, \(h = 0\)). The final mechanical energy \(E_f=K_{trans - pulley}+K_{rot - pulley}+K_{trans - spool}+K_{rot - spool}-mgh\)

The moment of inertia of a solid disk \(I=\frac{1}{2}mr^{2}\)

The translational kinetic energy of the pulley \(K_{trans - pulley}=\frac{1}{2}mv^{2}\), the rotational kinetic energy of the pulley \(K_{rot - pulley}=\frac{1}{2}I\omega^{2}=\frac{1}{2}(\frac{1}{2}mr^{2})(\frac{v}{r})^{2}=\frac{1}{4}mv^{2}\)

The translational kinetic energy of the spool \(K_{trans - spool}=\frac{1}{2}m(\frac{v}{2})^{2}=\frac{1}{8}mv^{2}\), the rotational kinetic energy of the spool \(K_{rot - spool}=\frac{1}{2}I\omega_s^{2}=\frac{1}{2}(\frac{1}{2}mr^{2})(\frac{v}{2r})^{2}=\frac{1}{16}mv^{2}\)

The potential energy change \(\Delta U=-mgh\)

By conservation of energy \(E_i = E_f\), so \(0=\frac{1}{2}mv^{2}+\frac{1}{4}mv^{2}+\frac{1}{8}mv^{2}+\frac{1}{16}mv^{2}-mgh\)

Cancel out \(m\) (since \(m
eq0\)):

$$
LATEXBLOCK0
$$

Given \(h = 2.25m\) and \(g = 9.8m/s^{2}\)

$$ v^{2}=\frac{16gh}{15} $$

$$ v=\sqrt{\frac{16\times9.8\times2.25}{15}} $$

$$ v=\sqrt{\frac{16\times9.8\times2.25}{15}}=\sqrt{\frac{16\times22.05}{15}}=\sqrt{\frac{352.8}{15}}\approx\sqrt{23.52}\approx4.85 $$

Answer:

\(4.85\)