Rotational Dynamics (AQA A Level Physics)

There is an analogy between quantities in rotational and translational dynamics.

Complete Table 1, stating in words the quantities in rotational dynamics that are analogous to force and mass in translational dynamics.

Table 1

Figure 1 shows a side view of the jib of a tower crane. The load is supported by a trolley which can move along the jib. The jib consists of all the parts of the crane above the bearing, but excluding the trolley and load.

Figure 1

The moment of inertia of the jib about the axis of rotation = 2.6 × 10⁷ kg m²

Mass of trolley and load = 2.2 × 10³ kg

The load is at a distance of 35 m from the axis of rotation.

Show that the total moment of inertia of the jib, and the trolley and load, about the axis of rotation is about 3 × 10⁷ kg m².

Figure 2 shows the variation of angular speed of the jib as it turns through an angle of 4.7 rad (270°) in a total time of 95 s. The trolley and load remain at a distance of 35 m from the axis.

Figure 2

Calculate the maximum angular speed of the jib.

At time X in Figure 2 the motor that is driving the jib is disengaged. A constant braking torque is then applied to bring the jib to a standstill from its maximum angular speed.

The crane driver repeats the movement of the jib with the same load at 35 m from the axis of rotation. Up to time X the motion is the same as before. From time X the trolley is driven at a steady speed away from the axis as the jib continues to rotate until the jib comes to a standstill.

Assume the braking torque remains the same as before.

Discuss how the motion of the trolley affects the time taken for the jib to come to a standstill.

Figure 2 shows a yo-yo made of two discs separated by a cylindrical axle. Thin string is wrapped tightly around the axle.

Figure 2

Initially both the free end A of the string and the yo-yo are held stationary.

With A remaining stationary, the yo-yo is now released so that it falls vertically. As the yo-yo falls, the string unwinds from the axle so that the yo-yo spins about its centre of mass.

The linear velocity of the centre of mass of the falling yo-yo is related to the angular velocity by where is the radius of the axle.

The yo-yo accelerates uniformly as it falls from rest. The string remains taut and has negligible thickness.

• mass of yo-yo = 9.2 × 10⁻² kg

• radius of axle = 5.0 × 10⁻³ m

• moment of inertia of yo-yo about axis X-X = 8.6 × 10⁻⁵ kg m²

When the yo-yo has fallen a distance of 0.50 m, its linear velocity is .

Calculate by considering the energy transfers that occur during the fall.

The yo-yo falls further until all the string is unwound. The yo-yo then ‘sleeps’. This means the yo-yo continues to rotate in a loose loop of string as shown in Figure 3.

Figure 3

The string applies a constant frictional torque of 8.3 × 10⁻⁴ N m to the axle.

The angular velocity of the yo-yo at the start of the sleep is 145 rad s⁻¹.

Determine, in rad, the total angle turned through by the yo-yo during the first 10 s of sleeping.

One equation used in translational dynamics is:

force = mass × acceleration

State in words the equivalent equation used in rotational dynamics.

Figure 1 shows two identical uniform plates A and B. The axis of rotation of each plate is shown.

Figure 1

State and explain which plate has the greater moment of inertia about its axis of rotation.

An electric motor drives a machine that punches out plates from a long strip of sheet metal. The motor runs continuously and is fitted with a flywheel.

Figure 2 shows how the angular velocity of the flywheel varies with time .

Figure 2

Table 1 describes the sequence for the machine after it has been brought up to speed .

Table 1

A new flywheel with a greater moment of inertia is fitted in place of the original flywheel. The motor torque is constant and the same as before.

Sketch on Figure 2 a graph showing how the angular velocity varies with time for the machine fitted with the new flywheel.

Assume that:

• the punching operation starts at the same angular speed

• the same quantity of energy is transferred when punching the metal sheet.

Explain one difference between your graph and the original graph.

Figure 3 shows a heavy stone grinding wheel used for sharpening tools.

Figure 3

The pedal is connected to the axle of the wheel by a connecting link and crank.

The operator pushes the pedal downwards to accelerate the wheel from rest.

The wheel begins to rotate in the direction shown.

Explain why the torque applied to the axle varies as the operator pushes downwards on the pedal.

The wheel is rotating at a high angular speed. The operator is told not to use the pedal to stop the rotation of the wheel suddenly.

Explain, with reference to angular impulse, why a sudden stop is likely to damage the mechanism.

The connecting link breaks. At this instant the angular speed of the wheel is 13.8 rad s⁻¹.

It takes 15.0 s for the wheel to come to rest.

The frictional torque acting at the axle bearings is 0.77 N m and is constant for all speeds.

Calculate the moment of inertia of the wheel.

The arrangement is modified as shown in Figure 4. The pedal, connecting link and crank are removed and the grinding wheel is driven by an electric motor.

Figure 4

Figure 4 also shows a tool being sharpened by pressing it on the edge of the rotating wheel.

The tool applies a tangential force on the wheel.

A torque of 3.10 N m is needed at the axle to drive the wheel at constant angular speed while the tool is being sharpened.

frictional torque at the axle bearings = 0.77 N m

radius of wheel = 0.24 m

Calculate .

Three motors E, F and G were available to drive the wheel in Figure 4.

Figure 5 shows how the torque varies with angular speed for each motor.

Figure 5

The no-load speed is the angular speed of a motor when the torque applied is zero.

The maximum power of each motor is developed at about .

The required output power of the motor when a tool is being sharpened is 52 W.

The required output power of the chosen motor should be about of its maximum power.

Deduce which motor E, F or G was chosen for this application.