IB Physics HL A.4 Rigid Body Mechanics

[Maximum number: 9]

A student models a rotating dancer using a system that consists of a vertical cylinder, a horizontal rod and two spheres.

The cylinder rotates from rest about the central vertical axis. A rod passes through the cylinder with a sphere on each side of the cylinder. Each sphere can move along the rod. Initially the spheres are close to the cylinder.

A horizontal force of 50 N is applied perpendicular to the rod at a distance of 0.50 m from the central axis. Another horizontal force of 40 N is applied in the opposite direction at a distance of 0.20 m from the central axis. Air resistance is negligible.

(a)

Show that the net torque on the system about the central axis is approximately 30 Nm .

[ 1 ]

$\Sigma \Gamma=50 \times 0.5+40 \times 0.2$

OR

33 «Nm»

Marking guidance:

Accept opposite rotational sign convention

(b)

The system rotates from rest and reaches a maximum angular speed of $20 \mathrm{rads}^{-1}$ in a time of 5.0 s . Calculate the angular acceleration of the system.

[ 1 ]

« $\mathrm{a}=\frac{20}{5}=$ » 4 «rad s $\mathrm{s}^{-2}$ »

(c)

Determine the moment of inertia of the system about the central axis.

[ 2 ]

$\begin{aligned} I=\frac{\Gamma}{\alpha} \text { OR } 33=I \times 4 I=8.25<\mathrm{kg} \mathrm{~m}^{2}> \end{aligned}$

Allow ECF from (a) and (b)

Award [2] for a BCA

(d)

When the system has reached its maximum angular speed, the two forces are removed. The spheres now move outward, away from the central axis.

[ 5 ]

(i)

Outline why the angular speed $\omega$ decreases when the spheres move outward.

[ 2 ]

moment of inertia increases

Angular momentum is conserved

Marking guidance:

Allow algebraic expressions e.g. $\omega=\frac{L}{I}$ so $\omega$ decreases for MP2

(ii)

Show that the rotational kinetic energy is $\frac{1}{2} L \omega$ where L is the angular momentum
of the system.

[ 1 ]

$E_{\kappa} \ll=\frac{1}{2} I \omega^{2}=>\frac{1}{2}(I \omega) \omega=\frac{1}{2} L \omega \vee$

Marking guidance:

Accept equivalent methods

(iii)

When the spheres move outward, the angular speed decreases from $20 \mathrm{rad} \mathrm{s}^{-1}$ to $12 \mathrm{rads}^{-1}$ . Calculate the percentage change in rotational kinetic energy that occurs when the spheres move outward.

[ 2 ]

$\begin{aligned} & \boldsymbol{«} \mathrm{E}_{\mathrm{k}}=» \frac{1}{2} L \omega_{1}=1 / 2 L \omega_{2} & \boldsymbol{O} \boldsymbol{R} & \frac{E_{k 1}}{E_{k 2}}=\frac{\omega_{1}}{\omega_{2}} \end{aligned}$

OR

« L is constant so» $E_{\mathrm{k}}$ is proportional to $\omega$

40 \% «energy loss»

MP1 is for understanding that angular momentum is constant so change in rotational kinetic energy is proportional to change in angular velocity

Award [0] if $E=0.5 \mathrm{I} \omega^{2}$ is used with the same I value for both values of E

Award [2] for BCA

[Maximum number: 6]

A uniform cylinder, of mass M and length L, has a moment of inertia of $\frac{1}{12} M L^{2}$ when rotated about an axis through its centre.

(a)

Outline what is meant by moment of inertia.

[ 1 ]

quotes $\mathrm{I}=\Sigma \mathrm{mr}^{2}$ with r= «perpendicular» distance to axis

OR

resistance to change in rotation

OR

ratio of torque «applied» to angular acceleration

OR

analog to mass in rotational mechanics

In MP1, accept r= radius.

In MP2, do not accept resistance to rotation.

In MP3 accept the expression as a formula if both symbols identified.

(b)

State the condition for rotational equilibrium.

[ 1 ]

(c)

Two identical cylinders, each of mass M and length L, are connected end to end. Show that the moment of inertia when these cylinders are rotated about their combined centre is $\frac{2}{3} M L^{2}$ .

[ 1 ]

$\frac{1}{12} 2 M(2 L)^{2}=<\frac{8}{12} M L^{2}=\frac{2}{3} M L^{2}>\vee$

(d)

A two-blade propeller can be modelled using the two-cylinder arrangement in (a)(iii).

The following data for the two-blade propeller are available:
Length of each blade: 0.60 m
Mass of each blade: 2.2 kg
Show that the moment of inertia of the two-blade propeller is about $0.5 \mathrm{~kg} \mathrm{~m}^{2}$ .

[ 1 ]

$\frac{2}{3} \times 2.2 \times 0.6^{2}$

OR
$\frac{1}{12} \times 4.4 \times 1.2^{2}$

OR

0.53 «kg $\mathrm{m}^{2}$ »

Answer of $0.5 \mathrm{~kg} \mathrm{~m}^{2}$ given, so award the
mark if candidates show a correct full substitution OR the value with an extra significant figure

(e)

The two-blade propeller is initially at rest. When a constant torque of 140 Nm acts on the two-blade propeller it reaches an angular speed of $750 \mathrm{rads}^{-1}$ . Ignore any frictional torque.

[ 2 ]

(i)

Calculate the time taken for the two-blade propeller to reach the angular speed of $750 \mathrm{rad} \mathrm{s}^{-1}$ .

[ 2 ]

«angular acceleration $=\frac{\text { torque }}{I}=$ »

Allow ECF from MP1

Award [2] for a BCA

Figure for Answer to Question (i) — IB Physics HL

[Maximum number: 5]

A boat is moved from land to water by rolling it across a set of cylindrical airbags.

(a)

Another way to move a boat down a ramp is to place it on rigid rods. Assume that the same boat is now moved on rods of diameter 1.80 m with the same boat acceleration as before.

[ 3 ]

(i)

Show that the angular displacement of a rod during the first 8.0 s of the motion is about 8 rad.

[ 1 ]

$\theta=2 \pi$ revolutions $=2 \pi \frac{s}{2 \pi r}=\frac{7}{0.9}=7.78 \mathrm{rad}$

(ii)

Outline how two frictional forces lead to the angular acceleration of the rod.

[ 2 ]

(b)

During the testing of the rod, it is rotated about its central axis. Point P is on the surface of the rod. At time t=0, P is at a point $45^{\circ}$ above the horizontal. The linear speed of P is $2.0 \mathrm{~ms}^{-1}$ .

The vertical displacement x of P can be modelled with an equation $x=x_{0} \sin (\omega t+\phi)$ .

[ 2 ]

(i)

Calculate the angular velocity $\omega$ of P.

[ 1 ]

$\omega=\frac{v}{r}=\frac{2}{0.9}=2.22 \mathrm{rads}^{-1}$

(ii)

Calculate the time it will take for P to reach the top of the circle.

[ 1 ]

Alternative 1
$t=\frac{\left(\frac{\mathrm{C}}{8}\right)}{v}=\frac{\left(\frac{2 \pi \mathrm{r}}{8}\right)}{2.0}=0.35 \mathrm{~s}$

Alternative 2
$t=\frac{\left(\frac{\pi}{4}\right)}{\omega}=0.35 \mathrm{~s}$

Allow 0.35-0.36 s.

[Maximum number: 1]

A student models a rotating dancer using a system that consists of a vertical cylinder, a horizontal rod and two spheres.

The cylinder rotates from rest about the central vertical axis. A rod passes through the cylinder with a sphere on each side of the cylinder. Each sphere can move along the rod. Initially the spheres are close to the cylinder.

A horizontal force of 50 N is applied perpendicular to the rod at a distance of 0.50 m from the central axis. Another horizontal force of 40 N is applied in the opposite direction at a distance of 0.20 m from the central axis. Air resistance is negligible.

Show that the net torque on the system about the central axis is approximately 30 Nm .

$\Sigma \Gamma=50 \times 0.5+40 \times 0.2$

OR

33 «Nm»

Marking guidance:

Accept opposite rotational sign convention

[Maximum number: 1]

A uniform cylinder, of mass M and length L, has a moment of inertia of $\frac{1}{12} M L^{2}$ when rotated about an axis through its centre.

State the condition for rotational equilibrium.

[Maximum number: 2]

A boat is moved from land to water by rolling it across a set of cylindrical airbags.

Another way to move a boat down a ramp is to place it on rigid rods. Assume that the same boat is now moved on rods of diameter 1.80 m with the same boat acceleration as before.

Outline how two frictional forces lead to the angular acceleration of the rod.

[Maximum number: 3]

A horizontal rigid bar of length 2 R is pivoted at its centre. The bar is free to rotate in a horizontal plane about a vertical axis through the pivot. A point particle of mass M is attached to one end of the bar and a container is attached to the other end of the bar.

A point particle of mass $\frac{M}{3}$ moving with speed v at right angles to the rod collides with the container and gets stuck in the container. The system then starts to rotate about the vertical axis.

The mass of the rod and the container can be neglected.

A torque of 0.010 Nm brings the system to rest after a number of revolutions. For this case $R=0.50 \mathrm{~m}, M=0.70 \mathrm{~kg}$ and $v=2.1 \mathrm{~ms}^{-1}$ .

Calculate the number of revolutions made by the system before it comes to rest.

$\theta=\frac{\omega_i^{2}}{2\alpha}$ , from $\omega_f^{2}=\omega_i^{2}+2\alpha\theta$ $\theta=\frac{v^{2}}{32R^{2}\alpha}=\frac{2.1^{2}}{32\times0.5^{2}\times0.043}=12.8\text{ or }12.9\ \mathrm{rad}$
number of rotations $=\frac{12.9}{2\pi}=2.0$ revolutions

[Maximum number: 1]

A horizontal rigid bar of length 2 R is pivoted at its centre. The bar is free to rotate in a horizontal plane about a vertical axis through the pivot. A point particle of mass M is attached to one end of the bar and a container is attached to the other end of the bar.

A point particle of mass $\frac{M}{3}$ moving with speed v at right angles to the rod collides with the container and gets stuck in the container. The system then starts to rotate about the vertical axis.

The mass of the rod and the container can be neglected.

Just after the collision the system begins to rotate about the vertical axis with angular velocity $\omega$ . Show that the angular momentum of the system is equal to $\frac{4}{3} M R^{2} \omega$ .

evidence of use of $L=I\omega=\left(MR^{2}+\frac{M}{3}R^{2}\right)\omega=\frac{4}{3}MR^{2}\omega$

evidence of use of conservation of angular momentum, $\frac{MvR}{3}=\frac{4}{3}MR^{2}\omega$ , rearranging to get $\omega=\frac{v}{4R}$

[Maximum number: 2]

A student models a rotating dancer using a system that consists of a vertical cylinder, a horizontal rod and two spheres.

The cylinder rotates from rest about the central vertical axis. A rod passes through the cylinder with a sphere on each side of the cylinder. Each sphere can move along the rod. Initially the spheres are close to the cylinder.

A horizontal force of 50 N is applied perpendicular to the rod at a distance of 0.50 m from the central axis. Another horizontal force of 40 N is applied in the opposite direction at a distance of 0.20 m from the central axis. Air resistance is negligible.

Determine the moment of inertia of the system about the central axis.

$\begin{aligned} I=\frac{\Gamma}{\alpha} \text { OR } 33=I \times 4 I=8.25<\mathrm{kg} \mathrm{~m}^{2}> \end{aligned}$

Allow ECF from (a) and (b)

Award [2] for a BCA

[Maximum number: 1]

A horizontal rigid bar of length 2 R is pivoted at its centre. The bar is free to rotate in a horizontal plane about a vertical axis through the pivot. A point particle of mass M is attached to one end of the bar and a container is attached to the other end of the bar.

A point particle of mass $\frac{M}{3}$ moving with speed v at right angles to the rod collides with the container and gets stuck in the container. The system then starts to rotate about the vertical axis.

The mass of the rod and the container can be neglected.

Write down an expression, in terms of M, v and R, for the angular momentum of the system about the vertical axis just before the collision.

$\frac{M}{3} v R$

[Maximum number: 1]

What is the unit of angular impulse?

A

Ns

B

Nm

C

$\mathrm{Nms}^{-1}$

D

Nms

[Maximum number: 1]

A student models a rotating dancer using a system that consists of a vertical cylinder, a horizontal rod and two spheres.

The cylinder rotates from rest about the central vertical axis. A rod passes through the cylinder with a sphere on each side of the cylinder. Each sphere can move along the rod. Initially the spheres are close to the cylinder.

A horizontal force of 50 N is applied perpendicular to the rod at a distance of 0.50 m from the central axis. Another horizontal force of 40 N is applied in the opposite direction at a distance of 0.20 m from the central axis. Air resistance is negligible.

When the system has reached its maximum angular speed, the two forces are removed. The spheres now move outward, away from the central axis.

Show that the rotational kinetic energy is $\frac{1}{2} L \omega$ where L is the angular momentum
of the system.

$E_{\kappa} \ll=\frac{1}{2} I \omega^{2}=>\frac{1}{2}(I \omega) \omega=\frac{1}{2} L \omega \vee$

Marking guidance:

Accept equivalent methods

IB Physics HL A 4 Rigid Body Mechanics Questions

Exam points

Covered syllabus points

Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question (d)

Question (i)

Question (ii)

Question (iii)

Question 2

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question (i)

Question 3

Question (a)

Question (i)

Question (ii)

Question (b)

Question (i)

Question (ii)

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12