TOPIC D: ROTATION ANSWERS SPRING 2018personalpages.manchester.ac.uk/staff/david.d.apsley/lectures/...The sphere is rolling, so that the total kinetic energy is the sum of: KE of linear

Mechanics Answers to Examples D (Rotation) - 1 David Apsley

TOPIC D: ROTATION – ANSWERS SPRING 2019

(Full worked answers follow on later pages)

Q1. 12.3 rad s–2

Q2. 17.5 rad s–1

Q3. (a)

(b)

(c)

(d)

Q4. (a) 15.2 N

(b) 3.91 m

(c) 11.8 m s–1

(d) 20.2 s

Q5. (a) 11.2 m s–1

(b) 14.5 kN

Q6. (a) 0.0028 kg m2

(b) 0.0035 kg m2

(c) 5.1 kg m2

Q7. (a) 0.0500 J

(b) 56 J

Q8. 63.7 N m

Q9. 7.67 rad s–1

Q10. (a) 3.10 m s–1

(b) 5.37 m s–1

Q11. (a) 32.6 kg m2

(b) v1 = 0.8ω; v2 = 0.2ω

(c) (i) the 60 kg mass; (ii) 4.05 rad s–1

; (iii) 3.34 m; 13.4 m

Q12. (a) 1.51109 kg m

2

(b) 1.58107 kg m

2 s

–1; 8.2910

4 J

(c) 52.8 kN

Q13. (a) 111 kg

(b) 9.43 kg m2; 0.292 m

(c) 436 J; 9.61 rad s–1

(d) 23.1 rad s–2


Q14. (a) (i) 3.67 m2

(ii) (across,down) = (0.8, 1.72) m, relative to corner A

(iii)2.14 m

(b) 0.870 rad s–1

Q15. (a) 71400 kg m2

(b) (i)

(ii) 0.224 rad s–2

; 2.47 rad s–1

(c) 37.1 m s–1

; 218 kJ

Q16. 0.35 m

Q17. (a)

(b)

Q18. (a)

(b)

(c)

Q19. (a) 5.4 kg m2

(b) 10.5 J

(c) 1.97 rad s–1

Q20. (a) 3.56 kg

(b) 0.537 m

(c) 1.62 kg m2

(d) 3.90 rad s–1

Q21. (a) 110 kg m2

(b) 2.03 m s–2

(c) 2.66 rad s–1

Q22. (a) 0.75 kg m2

(b) 29.0 J

(c) 0.0341 m below wheel centre

(d) 8.80 rad s–1

; 5.50 rad s–1

Q23.

,

,

,

Q24. No numerical answer

Q25. 65.4 rad s–1


Question 1.

Given:

initial velocity: u = 0

final velocity:

Hence,

linear acceleration

Since v = rω and r is constant, linear acceleration a and angular acceleration α are related by

Answer: 12.3 rad s–2

.


Question 2.


(167 rpm).


Question 3.

(a) The height at arbitrary point P is (height of the centre + height above centre) or

whilst the initial height is 2r.

KE + PE = constant

Answer: .

(b) Resolving radially (inward) with normal reaction force R pushing inward in this example:

Using the expression for v2 from part (a):

Answer: .

(c) The particle leaves the track when R = 0; i.e. at .

The speed at this point is given by

The vertical component of velocity at this point is

where

Hence,


Answer:

.

(d) The height when leaving the track is

The additional rise in height after leaving the track is determined by motion under gravity.

Using “ ” in the upward y direction, with initial velocity

and

acceleration –g, then, at the point of final rise (zero vertical velocity):

whence the further rise in height after leaving the track is

The total height is thus

Answer:

.


Question 4. (a) Resolving vertically:

Answer: 15.2 N.

(b) Find the extension of the rope from the tension:

The total length of the rope is then

Answer: 3.91 m.

(c) The speed of the object can be found from the centripetal acceleration, which, in turn, is

related to the horizontal component of the tension in the rope. Resolving inward:

The radius of the circle is

Answer: 11.8 m s–1

.

(d) Distance travelled in 10 complete circuits:

Time taken at speed v:


Alternative method: The angular velocity is

The time taken to rotate through angle 102π rad is then

Answer: 20.2 s.


Question 5. (a) The ball moves in an arc of a circle of radius r = 15 m. As the angle with the vertical

changes from 55º to 0º the ball drops a vertical distance

Gain in KE = loss in PE


.

(b) Equating the net upward force to mass centripetal acceleration:

Answer: 14.5 kN.


Question 6. (a) Model as a uniform plane disc, radius R = 0.15 m, mass M = 0.25 kg.

Answer: 0.0028 kg m2.

(b) Model as a spherical shell, radius R = circumference / (2π) = 0.1098 m, mass M = 0.43 kg.

Answer: 0.0035 kg m2.

(c) Model as a rectangular lamina (thickness is much smaller than any other dimension).

Mass:

The length perpendicular to the axis is the width of the door, L = 0.762 m.

Answer: 5.1 kg m2.


Question 7. (a) Solid cylinder, radius R = 0.15 m, mass M = 0.4 kg.

Angular velocity:

Kinetic energy (pure rotation):

Answer: 0.0500 J.

(b) Solid sphere, radius R = 0.05 m, mass M = 5 kg. Linear velocity of the centre of mass is

V = 4 m s–1

.

The sphere is rolling, so that the total kinetic energy is the sum of:

KE of linear motion of centre of mass

+

KE of rotational motion about the centre of mass.

To find the angular velocity from the linear velocity V note that the point of contact with the

floor is instantaneously at rest, so

Answer: 56 J.


Question 8.

For a rotating system, given torque T:

Answer: 63.7 N m.


Question 9. Gain in KE = loss in PE

(since the centre of mass has dropped a distance L/2). Hence,

For a uniform rod the moment of inertia about one end is

. Hence,


.


Question 10. Energy is conserved in the motion, so that

KE + PE = constant

Since the motion is rotational, write the kinetic energy in terms of ω (from which the velocity

at any radius can be found via v = rω). The potential energy depends on the heights of the

component parts above O, which can be written in terms of the angle θ made by the rod with

the upward vertical.

Kinetic Energy

The moment of inertia of the rod about its end at O is

The moment of inertia of the particle about O is

The total moment of inertia is

The kinetic energy is then given by

(ω in rad s

–1; KE in J)

Potential Energy

The particle is L cos θ above O, whilst the centre of mass of the rod is

above O.

Hence,

(in J).

Any other height datum would simply add a constant.

(a)


.

(b)



.


Question 11. (a) By summation the moment of inertia of flywheel + shaft is

Answer: 32.6 kg m2.

(b) Using “v = Rω”, since the velocity of each mass is equal

to the velocity of a particle on the perimeter:

where subscript 1 refers to flywheel and corresponding

cable, subscript 2 refers to shaft and corresponding cable.

Metre-second units are assumed in the numerical version.

(c)

(i) Energy statement:

KE + PE = constant

or

change in (KE + PE) = 0

Arbitrarily taking ω as positive anticlockwise and hence, for consistency, x1, v1 positive

downward, x2, v2 positive upward (but not making any assumption about whether each xi is

positive or negative):

Write velocities in terms of ω using part (b); write displacements in terms of x2, noting that

and hence .

Then,

Since the LHS is always positive then x2 must be positive. Hence the 60 kg mass (attached to

the shaft) rises and the 20 kg mass (attached to the flywheel) falls.

R

R

m

m

1

2

1

2

x1 v1, x2 v2,


(ii) Rearranging for ω, when x2 = 2 m,


.

(iii) 50 rpm corresponds to

Then

Answer: larger mass rises 3.34 m; smaller mass falls 13.4 m.


Question 12.

(a) The hoop has a mass of 1.2106 kg and all of this is at radius 30 m from the axis, so

(straight from its definition as second moment of mass, i.e. Σ(mass distance2)) its moment

of inertia is

Each carriage has mass 104 kg, all of which is 30 m from the axis, so its moment of inertia is

Each spoke has mass 3104 kg and length 30 m, so that, using the formula for the moment of

inertia of a rod, its moment of inertia is

spoke

For the combined assembly,

Answer: 1.51109 kg m

2.

(b) The rotation rate in radians per second is

Angular momentum:

Kinetic energy:

Answer: angular momentum = 1.58107 kg m

2 s

–1; kinetic energy = 8.2910

4 J.

(c) Angular-momentum principle:

torque = rate of change of angular momentum

Answer: retarding force 52.8 kN.


Question 13. (a)

Answer: 111 kg.

(b) The moment of inertia about the symmetry axis is

By the parallel-axis rule the moment of inertia about an axis a distance d (= 0.2 m) from an

axis through the centre of mass is

The radius of gyration k is defined by

Answer: moment of inertia = 9.43 kg m2; radius of gyration = 0.292 m.

(c) The maximum KE occurs where the PE is a minimum; i.e. where the centre of the plate

has dropped from 0.2 m above the axis to 0.2 m below it. Hence

At this point the angular velocity is also a maximum and is given by

Answer: KEmax = 436 J; ωmax = 9.61 rad s–1

.


(d) The maximum angular acceleration occurs where there is maximum torque. The torque is

caused by the weight of the plate and this is when the plate has turned 90°. The line of action

is then a distance d (= 0.2 m) from the axis. Hence:

Answer: maximum angular acceleration = 23.1 rad s–2

.


Question 14. Take coordinates with origin at A, x horizontal and y vertically downward. For a uniform

plane lamina the centre of mass coincides with the centre of area. Work in metres throughout.

(a)(i) The shape can be formed by subtraction of a circle from a rectangle. Subtracted areas

are marked negative so that they can be put in a formula with a Σ (sum).

Part 1 (rectangle):

Part 2 (circle):

Then

Answer: 3.67 m2.

(a)(ii) By symmetry,

whilst

Answer: (across,down) = (0.8, 1.72) m, relative to corner A.

(a)(iii) Find the moment of inertia about A by subtraction (rectangle – circle), then use the

definition: . All masses are simply ρarea, where ρ is the area mass density.

Rectangle:

This has mass

and the distance between the centroidal axis and A is

Hence, by the parallel-axis theorem and the given formula for its centre-of-mass axis:


Circle:

This has mass

and the distance between the centroidal axis and A is

Hence, by the parallel-axis theorem and the given formula for its centre-of-mass axis:

The net moment of inertia is then, by subtraction:

But

Hence,

k = 2.143 m

Answer: 2.14 m.

(b)

Gain in (rotational) KE = loss in (gravitational) PE

The centre of mass started a distance below A and swings round in an arc of a

circle. The largest KE, and hence largest angular velocity, occurs when PE is smallest; i.e.

when the centre of mass is lowest, directly below A, a distance (by Pythagoras’ Theorem) of

The distance dropped, which determines the loss in PE, is then

h = 1.893 – 1.716 = 0.177 m

From the energy statement above,

ω = 0.8696 rad s–1


.


Question 15. (a) Use the parallel-axis theorem to move the axis for the rods from their centres by distance

Summing the moments of inertia of cylinder and rod:

Answer: 71400 kg m2.

(b)(i)

(b)(ii)

Initial angular acceleration when ω = 0:

Final angular velocity when angular acceleration is 0:

Obviously, only the latter is possible.


; 2.47 rad s–1

.

(c)


; 218 kJ


Question 16. Method: first find the moment of inertia about an axis across a diameter. Then use the

parallel-axis theorem.

Let ρ be the density per unit area. By subtracting an inner disk of mass Mi and radius Ri from

an outer disk of mass Mo and radius Ro:

Also, for the total mass, by subtraction:

Hence,

By the parallel-axis theorem,

Hence,

Answer: 0.35 m.


Question 17. (a) Break down into small rectangles, dimensions 2y × δx. Let ρ be the area mass density.

Infinitesimal contributions to mass and moment of inertia are

Sum to obtain the total mass M and moment of inertia I and use these to eliminate ρ.

Mass:

(using symmetry and the shape of the lamina).

Substitute . Then

But

. Hence,

Moment of inertia:

Substitute . Then

But

Hence,


Finally, eliminate ρ:

Answer:

.

(b) Break down into small cylinders, radius y and height δx. Let ρ be the density.

Infinitesimal contributions to mass and moment of inertia are

Sum to obtain the total mass M and moment of inertia I and use these to eliminate ρ.

Finally, eliminate ρ:


Answer:

.


Question 18.

(a) The triangle has height

.

Using moments of length (since it has uniform 1-d elements), with lengths L having

individual centres of mass ½h, ½h and h below point A,

(There are symmetry arguments that could produce the same result.)

Answer: .

(b) The frame comprises two rods pivoted about their ends and one rod pivoted a distance

from its own centre of mass. Hence, using the formulae for moment of inertia of a rod

about its end or its centre, and the parallel-axis theorem for the bottom rod:

Answer:

.

(c) By conservation of energy:

gain in KE = loss in GPE

Answer:


Question 19.

(a) Model the gate as a rectangular lamina, mass M = 20 kg, dimension perpendicular to the

rotation axis L = 0.9 m. For an axis about one side,

Answer: 5.4 kg m2.

(b) To open the gate a moment (aka couple) must be applied to oppose the closing

mechanism. The work done by the opening moment (which equals the potential energy stored

in the springs of the gate) is

Answer: 10.5 J.

(c)



.


Question 20. (a)

Rectangle (1):

Semicircle (2):

Total mass:

Answer: 3.56 kg.

(b) One can answer this by finding either the centre of mass or centre of area. As the lamina

is uniform they will give the same answer. Here we use the first. Let x be the distance of

individual centres of mass from the axis, measured downward from O.

Rectangle:

Semicircle:

Overall centre of mass:

Answer: 0.537 m.

(c) Find the rectangle and semi-circle’s individual moments of inertia about their centres of

mass, then about the actual axis O. Then add to get the total moment of inertia about O. Note

that the semi-circle’s moment of inertia is first given (via the radius of gyration) about its

straight side and this must be shifted to its centre-of-mass axis before shifting again to O.

Rectangle:


Semi-circle:

Using the parallel-axis theorem in reverse to get to its centroid:

and the parallel-axis theorem again for the moment of inertia about the given axis through O:

The total moment of inertia is just the sum of the two parts:

Answer: 1.62 kg m2.

(d)



.


Question 21. (a) The overall moment of inertia can be found by subtracting the moments of inertia of the

small circular holes from that of the bigger circle. Formula for moment of inertia of a circle

(or circular cylinder) about an axis through its centre and perpendicular to the circle:

.

Large circle:

mass:

moment of inertia about O:

Each smaller circle:

mass:

moment of inertia about its centre:

and using the parallel-axis rule to move the axis to O:

moment of inertia about O:

Subtracting the four smaller circles:

Answer: 110 kg m2.

(b) Let F be the tension in the cable.

Linear motion of the mass:

(*)

Rotational motion of the wheel:

But ; hence:

(**)

Eliminating tension F by adding (*) and (**):

Hence,

R

F

F

mg

v



.

(c) Method 1: constant-acceleration

First find the speed of the mass by a constant-acceleration formula, and then use ω = v / R.

Method 2: energy

gain in KE = loss in PE

But v = rω. Hence,

(Any small differences are due to rounding the results of intermediate calculations.)


.


Question 22. (a)

Answer: 0.75 kg m2.

(b) Angular velocity:

Answer: 29.0 J.

(c) Taking the coordinate origin at the centre, with x horizontal and y vertically upward, the

depleted wheel can be constructed by subtracting one spoke from the entire wheel. So that it

can be used in a summation formula, the subtracted mass is marked negative. Work in kg and

m throughout.

Part 1 (intact wheel):

Part 2 (missing spoke)

At the moment of loss, by symmetry, whilst

Answer: 0.0341 m below the wheel centre.


(d) Immediately after losing a spoke, the depleted wheel is instantaneously still rotating at

8.796 rad s–1

. The new total mass and moment of inertia are (either from scratch or, as here,

by subtracting the contribution of one spoke):

The kinetic energy at this point is

As this is the lowest point of the centroid, it is the position with the minimum potential

energy and the maximum kinetic energy (and, hence, maximum angular velocity) in the

subsequent motion. So ωmax = 8.796 rad s–1

, its current rotation rate.

As the wheel rotates, the centroid moves above the wheel axis, the total rise being

which increases its potential energy by

The kinetic energy must decrease by the same amount, to the minimum value

The corresponding minimum angular velocity is given by

, or

Answer: ωmax = 8.80 rad s–1; ωmin = 5.50 rad s

–1.


Question 23. Let the mass of each rectangle be M; the mass of the whole lamina is 2M.

Consider first the axes perpendicular to the lamina. The moment of inertia of the upper

rectangle about its centre-of-mass axis A is (by formula)

Similarly,

The various moments of inertia of the whole lamina are then found by judicious application

of the parallel-axis rule for the individual rectangles and the radius of gyration by

(since the whole lamina has mass 2M)

NOTE: At no point is it necessary to find the centre of mass of the whole assembly. Just find

the moments of inertia of the individual rectangles and add them up!

Axis A

Axis B

Axis C


Axis D

For this axis in the plane of the lamina, both rectangles can be formed by stretching rods of

length L and L/3 respectively parallel to the axis, with moments of inertia about their centres.

Thus,

Axis E

One route (of many) is to work back from the perpendicular-axis rule:

Answer:

,

,

,

Equivalently,

(kA, kB, kC, kD, kE) = (0.561, 0.561, 0.948, 0.215, 0.518) L


Question 24. Consider first rolling motion and find the angular acceleration that would occur. This angular

acceleration is created by the frictional torque.

Resolving down the plane (F is friction):

(A)

Moments about centre of mass:

(B)

Eliminate F by r (A) + (B), and note that v = rω:

Hence, in rolling motion, the angular acceleration is given by:

(*)

The hoop will slip rather than roll if the maximum friction force is unable to provide the

torque necessary to accelerate it at this rate; i.e. if

Since and the angular acceleration is given by (*) slipping will

occur if

i.e.

For a hoop, . Hence the condition for slipping is

or


Question 25.

Doesn’t slip, so .

Resolving down the plane (F is friction):

(A)

Moments about centre of mass:

(B)

Eliminate F by r(A)+(B) and note that v = rω:

For a solid cylinder

Hence,

The angular acceleration is, therefore,

Since it starts from rest, its angular velocity after 5 s is


.

Documents

TOPIC D: ROTATION ANSWERS SPRING 2018personalpages.manchester.ac.uk/staff/david.d.apsley/lectures/...The sphere is rolling, so that the total kinetic energy is the sum of: KE of linear