6.01a Dimensions: M, L, T notation

One end of a rope is attached to a block A of mass 2 kg. The other end of the rope is attached to a second block B of mass 4 kg. Block A is held at rest on a fixed rough ramp inclined at \(30°\) to the horizontal. The rope is taut and passes over a small smooth pulley P which is fixed at the top of the ramp. The part of the rope from A to P is parallel to a line of greatest slope of the ramp. Block B hangs vertically below P, at a distance \(d\) m above the ground, as shown in the diagram. \includegraphics{figure_7} Block A is more than \(d\) m from P. The blocks are released from rest and A moves up the ramp. The coefficient of friction between A and the ramp is \(\frac{1}{2\sqrt{3}}\). The blocks are modelled as particles, the rope is modelled as light and inextensible, and air resistance can be ignored.

1. Determine, in terms of \(g\) and \(d\), the work done against friction as A moves \(d\) m up the ramp. [3]
2. Given that the speed of B immediately before it hits the ground is \(1.75 \text{ m s}^{-1}\), use the work–energy principle to determine the value of \(d\). [5]
3. Suggest one improvement, apart from including air resistance, that could be made to the model to make it more realistic. [1]

One end of a light spring is attached to a fixed point. A mass of 2 kg is attached to the other end of the spring. The spring hangs vertically in equilibrium. The extension of the spring is 0.05 m.

1. Find the stiffness of the spring. [2]
2. Find the energy stored in the spring. [2]
3. Find the dimensions of stiffness of a spring. [1]

A particle P of mass \(m\) is performing complete oscillations with amplitude \(a\) on the end of a light spring with stiffness \(k\). The spring hangs vertically and the maximum speed \(v\) of P is given by the formula $$v = Cm^{\alpha}a^{\beta}k^{\gamma},$$ where C is a dimensionless constant.

1. Use dimensional analysis to determine \(\alpha\), \(\beta\), and \(\gamma\). [4]

Fig. 6 shows a pendulum which consists of a rod AB freely hinged at the end A with a weight at the end B. The pendulum is oscillating in a vertical plane. The total energy, \(E\), of the pendulum is given by $$E = \frac{1}{2}I\omega^2 - mgh\cos\theta,$$ where

• \(\omega\) is its angular speed
• \(m\) is its mass
• \(h\) is the distance of its centre of mass from A
• \(\theta\) is the angle the rod makes with the downward vertical
• \(g\) is the acceleration due to gravity
• \(I\) is a quantity known as the moment of inertia of the pendulum.

\includegraphics{figure_6}

1. Use the expression for \(E\) to deduce the dimensions of \(I\). [4]

It is suggested that the period of oscillation, \(T\), of the pendulum is given by \(T = kI^\alpha(mg)^\beta h^\gamma\), where \(k\) is a dimensionless constant.

1. Use dimensional analysis to find the values of \(\alpha\), \(\beta\) and \(\gamma\). [5]

A class experiment finds that, when all other quantities are fixed, \(T\) is proportional to \(\frac{1}{\sqrt{m}}\).

1. Determine whether this result is consistent with your answer to part (ii). [1]

A particle of mass \(m\) moves in a straight line with constant acceleration \(a\). Its initial and final velocities are \(u\) and \(v\) respectively and its final displacement from its starting position is \(s\). In order to model the motion of the particle it is suggested that the velocity is given by the equation $$v^2 = pu^{\alpha} + qa^{\beta}s^{\gamma}$$ where \(p\) and \(q\) are dimensionless constants.

1. Explain why \(\alpha\) must equal 2 for the equation to be dimensionally consistent. [2]
2. By using dimensional analysis, determine the values of \(\beta\) and \(\gamma\). [4]
3. By considering the case where \(s = 0\), determine the value of \(p\). [1]
4. By multiplying both sides of the equation by \(\frac{1}{2}m\), and using the numerical values of \(\alpha\), \(\beta\) and \(\gamma\), determine the value of \(q\). [2]

One end of a non-uniform rod is freely hinged to a fixed point so that the rod can rotate about the point. When the rod rotates with angular velocity \(\omega\) it can be shown that the kinetic energy \(E\) of the rod is given by \(E = \frac{1}{2}I\omega^2\), where \(I\) is a quantity called the moment of inertia of the rod.

1. Deduce the dimensions of \(I\). [3]
2. Given that the rod has mass \(m\) and length \(r\), suggest an expression for \(I\), explaining any additional symbols that you use. [3]

A student notices that the formula \(E = \frac{1}{2}I\omega^2\) looks similar to the formula \(E = \frac{1}{2}mv^2\) for the kinetic energy of a particle, with angular velocity for the rod corresponding to velocity for the particle, and moment of inertia corresponding to mass. Assuming a similar correspondence between angular acceleration (i.e. \(\frac{d\omega}{dt}\)) and acceleration, the student thinks that an equation for angular motion of the rod corresponding to Newton's second law for the particle should be \(F = I\alpha\), where \(F\) is the force applied to the rod and \(\alpha\) is the resulting angular acceleration.

1. Use dimensional analysis to show that the student's suggestion is incorrect. [2]
2. State the dimensions of a quantity \(x\) for which the equation \(Fx = I\alpha\) would be dimensionally consistent. [1]
3. Explain why the fact that the equation in part (iv) is dimensionally consistent does not necessarily mean that it is correct. [1]