Question

Two masses \(m\) and \(2m\) are connected by a light string going over a pulley (disc) of mass \(30m\) with radius \(r=0.1\,\text{m}\). The pulley is mounted in a vertical plane and is free to rotate about its axis. The \(2m\) mass is released from rest and its speed when it has descended through a height of \(3.6\,\text{m}\) is ____________ m/s. (Assume string does not slip and \(g=10\,\text{m s}^{-2}\)).

Hint:

Always include rotational kinetic energy when pulleys have mass and the string does not slip.

Answer 1

Correct Answer: 4

To solve the problem, we'll apply the conservation of mechanical energy. Initially, the system has gravitational potential energy, which converts into kinetic energy of the masses and rotational energy of the pulley. The gravitational potential energy for mass \(2m\) descending height \(h=3.6\,\text{m}\) is: 

\(U=2mgh=2m \times 10 \times 3.6=72m\,\text{J}\)

The kinetic energies are:

• Translational kinetic energy of mass \(2m\): \(K_1=\frac{1}{2}(2m)v^2=mv^2\)
• Translational kinetic energy of mass \(m\): \(K_2=\frac{1}{2}m v^2\)
• Rotational kinetic energy of the pulley (moment of inertia \(I=\frac{1}{2}MR^2\)): \(K_3=\frac{1}{2}\left(\frac{1}{2}(30m)r^2\right)\omega^2=\frac{1}{4}(30m)r^2\omega^2=\frac{15mr^2\omega^2}{4}\)

Since the string doesn't slip, the linear speed \(v\) is related to angular speed \(\omega\) as \(v=r\omega\). By substituting \(\omega=\frac{v}{r}\), we have:

\(K_3=\frac{15mr^2}{4}\left(\frac{v}{r}\right)^2=\frac{15mv^2}{4}\)

Total energy:

\(mv^2+\frac{15mv^2}{4}+mv^2=U=72m\)

Simplifying:

\(mv^2+\frac{15mv^2}{4}+mv^2 =72m \rightarrow (1+\frac{15}{4}+1)mv^2=72m \rightarrow \frac{11mv^2}{2}=72m\)

Solving for \(v^2\):

\(11v^2=144 \rightarrow v^2=\frac{144}{11} \rightarrow v=\sqrt{\frac{144}{11}}\)

Calculating \(v\):

\(v\approx\sqrt{13.09}\approx3.619\,\text{m/s}\)

This result is within the expected range of 4,4. The speed of the \(2m\) mass after descending is approximately \(3.6\,\text{m/s}\).