Category: Machine Kinematics Objective Questions

250+ TOP MCQs on Kinematic Pair and Answers

Machine Kinematics Multiple Choice Questions on “Kinematic Pair”.

1. When the two elements of a pair have _____________ when in motion, it is said to a lower pair.
a) line or point contact
b) surface contact
c) permit relative motion
d) none of the mentioned
Answer: b
Clarification: When the two elements of a pair have surface contact when relative motion, takes place and the surface of one element slides over the surface of the other, the pair formed is known as lower pair. It will be seen that sliding pairs, turning pairs and screw pairs form lower pairs.

2. The two elements of a pair are said to form a higher pair, when they
a) have a surface contact when in motion
b) have a line or point contact when in motion
c) are kept in contact by the action of external forces, when in motion
d) permit relative motion
Answer: b
Clarification: When the two elements of a pair have a line or point contact when relative motion takes place and the motion between the two elements is partly turning and partly sliding, then the pair is known as higher pair.

3. In a force-closed pair, the two elements of a pair are not held together mechanically.
a) True
b) False
Answer: b
Clarification: When the two elements of pair are not connected mechanically but are kept in contact by the action of external forces, the pair is said to be a forced-closed pair.

4. The two elements of a pair are said to form a ___________ when they permit relative motion between them.
a) open pair
b) kinematic pair
c) higher pair
d) lower pair
Answer: b
Clarification: The two links or elements of a machine, when in contact with each other, are said to form a pair. If the relative motion between them is completely or successfully constrained, the pair is known as kinematic pair.

5. In an open pair, the two elements of a pair
a) have a surface contact when in motion
b) have a line or point contact when in motion
c) are kept in contact by the action of external forces, when in motion
d) are not held mechanically
Answer: d
Clarification: When the two elements of a pair are not held mechanically, they are called open pair.

6. The sliding pairs, turning pairs and screw pairs form lower pairs.
a) True
b) False
Answer: a
Clarification: When the two elements of a pair have surface contact when relative motion, takes place and the surface of one element slides over the surface of the other, the pair formed is known as lower pair. It will be seen that sliding pairs, turning pairs and screw pairs form lower pairs.

7. A combination of kinematic pairs, joined in such a way that the relative motion between the links is completely constrained, is called a
a) structure
b) mechanism
c) kinematic chain
d) inversion
Answer: c
Clarification: A kinematic chain is defined as a combination of kinematic pairs, joined in such a way that each link forms a part of two pairs and the relative motion between the links or elements is completely or successfully constrained.

8. The relation between number of pairs(p) forming a kinematic chain and the number of links(l) is
a) l = 2p – 2
b) l = 2p – 3
c) l = 2p – 4
d) l = 2p – 5
Answer: c
Clarification: If each link is assumed to form two pairs with adjacent links, then the relation between the number of pairs(p) forming a kinematic chain and the number of links(l) may be expressed in the form of an equation : l = 2p – 4

9. The relation between number of links(l) and number of joints(j) in a kinematic chain is
a) l = 1/2 (j+2)
b) l = 2/3 (j+2)
c) l = 3/4 (j+2)
d) l = j+4
Answer: b
Clarification: Another relation between the number of links (l) and the number of joints(j) which constitute a kinematic chain is given by the expression : l = 2/3 (j+2)

10. The relation l = 2/3(j+2) apply to kinematic chains in which lower pairs are used. This may be used to kinematic chains in which higher pairs are used, but each higher pair may be taken as equivalent to
a) one lower pair and two additional links
b) two lower pairs and one additional link
c) two lower pairs and two additional links
d) all of the mentioned
Answer: b
Clarification: None

250+ TOP MCQs on Properties of Instantaneous Centre and Answers

Machine Kinematics Multiple Choice Questions on “Properties of Instantaneous Centre”.

1. Which is the false statement about the properties of instantaneous centre?
a) at the instantaneous centre of rotation, one rigid link rotates instantaneously relative to another for the configuration of mechanism considered
b) the two rigid links have no linear velocities relative to each other at the instantaneous centre
c) the two rigid links which have no linear velocity relative to each other at this centre have the same linear velocity to the third rigid link
d) the double centre can be denoted either by O21 or O12, but proper selection should be made
Answer: d
Clarification: The following properties of the instantaneous centre are important from the subject point of view :
1. A rigid link rotates instantaneously relative to another link at the instantaneous centre for the configuration of the mechanism considered.
2. The two rigid links have no linear velocity relative to each other at the instantaneous centre. At this point (i.e. instantaneous centre), the two rigid links have the same linear velocity relative to the third rigid link. In other words, the velocity of the instantaneous centre relative to any third rigid link will be same whether the instantaneous centre is regarded as a point on the first rigid link or on the second rigid link.

2. Instantaneous center of rotation of a link in a four bar mechanism lies on
a) right side pivot of this link
b) left side pivot of this link
c) a point obtained by intersection on extending adjoining links
d) none of the mentioned
Answer: c
Clarification: None.

3. The total number of instantaneous centers for a mechanism of n links is
a) n(n – 1)/2
b) n
c) n – 1
d) n(n – 1)
Answer: a
Clarification: The number of pairs of links or the number of instantaneous centres is the number of combinations of n links taken two at a time. Mathematically, number of instantaneous centres,
N = n(n – 1)/2.

4. The number of links and instantaneous centers in a reciprocating engine mechanism are
a) 4,4
b) 4,5
c) 5,4
d) 4,6
Answer: d
Clarification: First of all, determine the number of instantaneous centres (N) by using the relation
N = n(n – 1)/2
In present case, N = 4(4 – 1)/2 (n = 4)
= 6.

5. According to Kennedy’s theorem, if three bodies have plane motions, their instantaneous centres lie on
a) a triangle
b) a point
c) two lines
d) a straight line
Answer: d
Clarification: The Aronhold Kennedy’s theorem states that if three bodies move relatively to each other, they have three instantaneous centres and lie on a straight line.

6. In a rigid link OA, velocity of A w.r.t. O will be
a) parallel to OA
b) perpendicular to OA
c) at 450 to OA
d) along AO
Answer: b
Clarification: None.

7. Two systems shall be dynamically equivalent when
a) the mass of two are same
b) c.g. of two coincides
c) M.I. of two about an axis through c.g. is equal
d) all of the mentioned
Answer: d
Clarification: None.

8. A link is rotating about O. Velocity of point P on link w.r.t. point Q on link will be perpendicular to
a) OP
b) OQ
c) PQ
d) line in between OP and OQ
Answer: c
Clarification: A link is rotating about O. Velocity of point P on link w.r.t. point Q on link will be perpendicular to PQ.
The velocity of any point in mechanism relative to any other point on the mechanism on velocity polygon is represented by the line joining the corresponding points.

9. The velocity of any point in mechanism relative to any other point on the mechanism on velocity polygon is represented by the line
a) joining the corresponding points
b) perpendicular to line
c) at 450 to line
d) none of the mentioned
Answer: a
Clarification: A link is rotating about O. Velocity of point P on link w.r.t. point Q on link will be perpendicular to PQ.
The velocity of any point in mechanism relative to any other point on the mechanism on velocity polygon is represented by the line joining the corresponding points.

10. The absolute acceleration of any point P in a link about center of rotation O is
a) along PO
b) perpendicular to PO
c) at 450 to PO
d) none of the mentioned
Answer: d
Clarification: The coriolis component of acceleration is always perpendicular to the link.

11. Angular acceleration of a link can be determined by dividing the
a) centripetal component of acceleration with length of link
b) tangential component of acceleration with length of link
c) resultant acceleration with length of link
d) all of the mentioned
Answer: b
Clarification: The angular acceleration of the link AB is obtained by dividing the tangential components of the acceleration of B with respect to A to the length of the link.

250+ TOP MCQs on Double Hooke’s Joint and Answers

Machine Kinematics Multiple Choice Questions on “Double Hooke’s Joint”.

1. What is the purpose of double hooke’s joint?
a) Have constant linear velocity ratio of driver and driven shafts
b) Have constant acceleration ratio of driver and driven shafts
c) Have constant angular velocity ratio of driver and driven shafts
d) Have constant angular acceleration ratio of driver and driven shafts
Answer: c
Clarification: The velocity of the driven shaft is not constant, but varies from maximum to minimum values. In order to have a constant velocity ratio of the driving and driven shafts, an intermediate shaft with a Hooke’s joint at each end is used.

2. Double hooke’s joint can be used to keep the angular velocity of the shaft constant.
a) True
b) False
Answer: b
Clarification: Double hooke’s joint is used to keep the velocity ratio of driver shaft and driven shaft, It does not necessarily keeps the velocity constant.

3. Two shafts having an included angle of 150° are connected by a Hooke’s joint. The driving shaft runs at a uniform speed of 1500 r.p.m. The driven shaft carries a flywheel of mass 12 kg and 100 mm radius of gyration. Using the above data, calculate the maximum angular acceleration of the driven shaft in rad/s2.
a) 6853
b) 6090
c) 6100
d) 6500
Answer: a
Clarification: α = 180 -150 = 30⁰
cos2θ = 2sin2 α/1-sin2 α = 0.66
angular acc = dω/dt
= 6853.0 rad/s2.

4. Two shafts having an included angle of 150° are connected by a Hooke’s joint. The driving shaft runs at a uniform speed of 1500 r.p.m. The driven shaft carries a flywheel of mass 12 kg and 100 mm radius of gyration. Using the above data, calculate the maximum torque required in N-m.
a) 822
b) 888
c) 890
d) 867
Answer: a
Clarification: α = 180 -160 = 30⁰
cos2θ = 2sin2 α/1-sin2 α = 0.66
angular acc = dω/dt
= 6853 rad/s2
I = 0.12 Kg-m2
Therefore max torque = I.ang acc.
= 822 N-m.

5. Two shafts connected by a Hooke’s joint have an angle of 18 degrees between the axes.
Find the angle through which it should be turned to get the velocity ratio maximum.
a) 180
b) 30
c) 45
d) 90
Answer: a
Clarification: Velocity ratio is ω1/ω = cosα/(1 – cos2θsin2α)
now this to be maximum cos2θ = 1
therefore θ = 0 or 180 degrees.

6. Two shafts connected by a Hooke’s joint have an angle of 18 degrees between the axes.
Find the angle through which it should be turned to get the velocity ratio equal to 1.
a) 30.6
b) 30.3
c) 44.3
d) 91.2
Answer: c
Clarification: Velocity ratio is ω1/ω = cosα/(1 – cos2θsin2α)
now this to be 1
we get, cosα = 1 – cos2θsin2α
solving this equation we get
θ = 44.3 or 135.7 degrees.

7. Two shafts with an included angle of 160° are connected by a Hooke’s joint. The driving shaft runs at a uniform speed of 1500 r.p.m. The driven shaft carries a flywheel of mass 12 kg and 100 mm radius of gyration. Find the maximum angular acceleration of the driven shaft.
a) 3090 rad/s2
b) 4090 rad/s2
c) 5090 rad/s2
d) 6090 rad/s2
Answer: a
Clarification: Given : α = 180° – 160° = 20°; N = 1500 r.p.m.; m = 12 kg ; k = 100 mm = 0.1 m
We know that angular speed of the driving shaft,
ω = 2 π × 1500 / 60 = 157 rad/s
and mass moment of inertia of the driven shaft,
I = m.k2 = 12(0.1)2 = 0.12 kg – m2

Let dω1 / dt = Maximum angular acceleration of the driven shaft, and
θ = Angle through which the driving shaft turns.
We know that, for maximum angular acceleration of the driven shaft,

cos 2θ = 2sin2α/2 – sin2α = 2sin220°/2 – sin220° = 0.124
2θ = 82.9° or θ = 41.45°
and dω1 / dt = ω2cosα sin2θsin2α/(1 – cos2θsin2α)2
= 3090 rad/s2.

8. The angle between the axes of two shafts connected by Hooke’s joint is 18°. Determine the angle turned through by the driving shaft when the velocity ratio is maximum.
a) 90°
b) 180°
c) 270°
d) 360°
Answer: b
Clarification: Given : α = 98°
Let θ = Angle turned through by the driving shaft.
We know that velocity ratio,
ω1/ω = cosα/1 – cos2θsin2α

The velocity ratio will be maximum when cos2 θ is minimum, i.e. when
cos2 θ = 1 or when θ = 0° or 180°.

contest

250+ TOP MCQs on Types of Belts – 1 and Answers

Machine Kinematics Multiple Choice Questions on “Types of Belts – 1”.

1. In a cone pulley, if the sum of radii of the pulleys on the driving and driven shafts is constant, then
a) open belt drive is recommended
b) crossed belt drive is recommended
c) both open belt drive and crossed belt drive is recommended
d) the drive is recommended depending upon the torque transmitted
Answer: b
Clarification: Cone pulley drive, is used for changing the speed of the driven shaft while the main or driving shaft runs at constant speed. This is accomplished by shifting the belt from one part of the steps to the other.

2. Due to slip of belt, the velocity ratio of the belt drive increases.
a) True
b) False
Answer: b
Clarification: The result of the belt slipping is to reduce the velocity ratio of the system.

3. When two pulleys of different diameters are connected by means of an open belt, the angle of contact at the _________pulley must be taken into consideration.
a) smaller
b) larger
c) medium
d) none of the mentioned
Answer: a
Clarification: None.

4. The power transmitted by a belt is maximum when the maximum tension in the belt is __________of centrifugal tension.
a) one-third
b) two-third
c) double
d) three times
Answer: d
Clarification: When the power transmitted is maximum, 1/3rd of the maximum tension is absorbed as centrifugal tension.

5. The velocity of the belt for maximum power is
a) T/3
b) T.g/3
c) √T/3m
d) √3m/T
Answer: c
Clarification: None.

6. The centrifugal tension on the belt has no effect on the power transmitted.
a) True
b) False
Answer: a
Clarification: None.

7. V-belts are usually used for
a) long drives
b) short drives
c) long and short drives
d) none of the mentioned
Answer: b
Clarification: V-belt is mostly used in factories and workshops where a great amount of power is to be transmitted from one pulley to another when the two pulleys are very near to each other.

8. In a multiple V-belt drive, if one of the belt is broken, then we should replace
a) the broken belt only
b) all the belts
c) the broken belt and the belts on either side of it
d) none of the mentioned
Answer: b
Clarification: In multiple V-belt drive, all the belts should stretch at the same rate so that the load is equally divided between them. When one of the set of belts break, the entire set should be replaced at the same time. If only one belt is replaced, the new unworn and unstressed belt will be more tightly stretched and will move with different velocity.

9. The included angle for the v-belt is usually
a) 100 to 200
b) 200 to 300
c) 300 to 400
d) 600 to 800
Answer: c
Clarification: None.

250+ TOP MCQs on Standard Proportions of Gear Systems and Answers

Machine Kinematics Multiple Choice Questions on “Standard Proportions of Gear Systems”.

1. If T is the actual number of teeth on a helical gear and φ is the helix angle for the teeth, the formative number of teeth is written as
a) T sec3 φ
b) T sec2 φ
c) T/sec3φ
d) T cosec φ

Answer: a
Clarification: The formative or equivalent number of teeth for a helical gear may be defined as the number of teeth that can be generated on the surface of a cylinder having a radius equal to the radius of curvature at a point at the tip of the minor axis of an ellipse obtained by taking a section of the gear in the normal plane. Mathematically, formative or equivalent number of teeth on a helical gear,T sec3 φ

2. In helical gears, the distance between similar faces of adjacent teeth along a helix on the pitch cylinders normal to the teeth, is called
a) normal pitch
b) axial pitch
c) diametral pitch
d) module

Answer: a
Clarification: Normal pitch is the distance between similar faces of adjacent teeth along a helix on the pitch cylinders normal to the teeth.
Axial pitch is the distance, parallel to the axis, between similar faces of adjacent teeth.

3. In helical gears, the right hand helices on one gear will mesh ____________ helices on the other gear.
a) right hand
b) left hand
c) opposite
d) none of the mentioned

Answer: b
Clarification: A helical gear has teeth in form of helix around the gear. Two such gears may be used to connect two parallel shafts in place of spur gears. The helixes may be right handed on one gear and left handed on the other.

4. The helix angle for single helical gears ranges from
a) 10° to 15°
b) 15° to 20°
c) 20° to 35°
d) 35° to 50°

Answer: c
Clarification: In single helical gears, the helix angle ranges from 20° to 35°, while for double helical gears, it may be made upto 45°.

5. The helix angle for double helical gears may be made up to
a) 45°
b) 60°
c) 75°
d) 90°

Answer: a
Clarification: In single helical gears, the helix angle ranges from 20° to 35°, while for double helical gears, it may be made upto 45°.

6. The outside diameter of an involute gear is equal to pitch circle diameter plus
a) 2 addendum
b) 2 dedendum
c) 3.1416 module
d) 2.157 module

Answer: a
Clarification: Addendum is the portion of gear tooth above the pitch circle diameter (PCD). Therefore outside diameter of involute gear = PCD + 2 addendum.

7. Pick out the false statement about relationships of spur gears.
a) Pitch diameter = module x No. of teeth
b) Module = 25.4/diametral pitch
c) dedendum = 1.25 x module
d) Base pitch = module x п x sinɸ

Answer: d
Clarification: The false statement is
Base pitch = module x п x sinɸ
Correct relationship of base pitch = module x п x cosɸ

8. Which of the following is not the correct property of involute curve?
a) The form or shape pf an involute curve depends upon the diameter of base circle from which it is derived
b) The angular motion of two involute gear teeth rotating at a uniform rate will be uniform, irrespective of the centre distance
c) The relative rate of motion between driving and driven gears having involute tooth curves, is established by the diameters of their pitch circles
d) the pitch diameters of mating involute gears are directly proportional to the diameters of their respective base circles

Answer: c
Clarification: All statements except at (c) are correct. The correct statement for (c) is – The relative rate motion between driving and driven gears having involute tooth curves, is established by the diameters of their base circles.

9. Which of the following gear ratio does not result in hunting tool
a) 77/20
b) 76/21
c) 75/22
d) 71/25

Answer: d
Clarification: When several pairs of gears operating at the same centre distance are required to have hunting ratios, this can be accomplished by having the sum of the teeth in each pair equal to a prime number.

10. In measuring the chordal thickness, the vertical scale of a gear tooth caliper is set to the chordal or corrected addendum to locate the caliper jaws at the pitch line.If a = addendum, t = circular thickness of tooth at pitch diameter D, then chordal thickness is equal to
a) a + t/D
b) a + t2/D
c) a + t3/2D
d) a + t2/4D

Answer: d
Clarification: The correct relationship is chordal thickness = a + t2/4D

250+ TOP MCQs on Numericals On Kinetics Of Motion and Loss of Kinetic Energy and Answers

Machine Kinematics Interview Questions on “Numericals On Kinetics Of Motion and Loss of Kinetic Energy”.

1. A road roller has a total mass of 12 tonnes. The front roller has a mass of 2 tonnes, a radius of gyration of 0.4 m and a diameter of 1.2 m. The rear axle, together with its wheels, has a mass of 2.5 tonnes, a radius of gyration of 0.6 m and a diameter of 1.5 m. Calculate kinetic energy of rotation of the wheels and axles at a speed of 9 km/h.
a) 7670 N-m
b) 8670 N-m
c) 9670 N-m
d) 6670 N-m

Answer: a
Clarification: Given : m = 12 t = 12 000 kg ;
m1 = 2 t = 2000 kg ; k1 = 0.4 m ; d1 = 1.2 m or r1 = 0.6 m ; m2 = 2.5 t = 2500 kg ; k2 = 0.6 m ; d2 = 1.5 m or r2 = 0.75 m ; v = 9 km/h = 2.5 m/s; s = 6 m

Kinetic energy of rotation of the wheels and axles

We know that mass moment of inertia of the front roller,
I1 = m1(k1)2 = 2000 (0.4)2 = 320 kg-m2

and mass moment of inertia of the rear axle together with its wheels,
I2 = m2 (k2)2 = 2500 (0.6)2 = 900 kg -m2

Angular speed of the front roller,
ω1 = v/r1 = 2.5/0.6 = 4.16 rad/s
and angular speed of rear wheels,

ω2 = v/r2 = 2.5/0.75 = 3.3 rad/s

We know that kinetic energy of rotation of the front roller,

E1 =1/2 I1 (ω1)2 = 1/2 × 320(4.16)2 2770 N-m

and kinetic energy of rotation of the rear axle together with its wheels,

E2 =1/2 I2 (ω2)2 = 1/2 × 900(3.3)2 4900 N-m

∴ Total kinetic energy of rotation of the wheels,
E = E1 + E2 = 2770 + 4900 = 7670 N-m

2. A road roller has a total mass of 12 tonnes. The front roller has a mass of 2 tonnes, a radius of gyration of 0.4 m and a diameter of 1.2 m. The rear axle, together with its wheels, has a mass of 2.5 tonnes, a radius of gyration of 0.6 m and a diameter of 1.5 m. Calculate total kinetic energy of road roller.
a) 25170 N-m
b) 35170 N-m
c) 45170 N-m
d) 55170 N-m

Answer: d
Clarification: Given : m = 12 t = 12 000 kg ;
m1 = 2 t = 2000 kg ; k1 = 0.4 m ; d1 = 1.2 m or r1 = 0.6 m ; m2 = 2.5 t = 2500 kg ; k2 = 0.6 m ; d2 = 1.5 m or r2 = 0.75 m ; v = 9 km/h = 2.5 m/s; s = 6 m

Kinetic energy of rotation of the wheels and axles

We know that mass moment of inertia of the front roller,
I1 = m1(k1)2 = 2000 (0.4)2 = 320 kg-m2

and mass moment of inertia of the rear axle together with its wheels,
I2 = m2 (k2)2 = 2500 (0.6)2 = 900 kg -m2

Angular speed of the front roller,
ω1 = v/r1 = 2.5/0.6 = 4.16 rad/s
and angular speed of rear wheels,

ω2 = v/r2 = 2.5/0.75 = 3.3 rad/s

We know that kinetic energy of rotation of the front roller,

E1 =1/2 I1 (ω1)2 = 1/2 × 320(4.16)2 2770 N-m

and kinetic energy of rotation of the rear axle together with its wheels,

E2 =1/2 I2 (ω2)2 = 1/2 × 900(3.3)2 4900 N-m

∴ Total kinetic energy of rotation of the wheels,
E = E1 + E2 = 2770 + 4900 = 7670 N-m

We know that the kinetic energy of motion (i.e. kinetic energy of translation) of the road roller,
E3 = 1/2 mv2 = 1/2 x 1200 (2.5)2 = 37500 N-m

This energy includes the kinetic energy of translation of the wheels also, because the total mass (m) has been considered.

∴ Total kinetic energy of road roller,
E4 = Kinetic energy of translation + Kinetic energy of rotation
= E3 + E = 37 500 + 7670 = 45 170 N-m

3. A road roller has a total mass of 12 tonnes. The front roller has a mass of 2 tonnes, a radius of gyration of 0.4 m and a diameter of 1.2 m. The rear axle, together with its wheels, has a mass of 2.5 tonnes, a radius of gyration of 0.6 m and a diameter of 1.5 m. Calculate braking force required to bring the roller to rest from 9 km/h in 6 m on the level.
a) 5528.3 N
b) 6528.3 N
c) 7528.3 N
d) 8528.3 N

Answer: c
Clarification: Given : m = 12 t = 12 000 kg ;
m1 = 2 t = 2000 kg ; k1 = 0.4 m ; d1 = 1.2 m or r1 = 0.6 m ; m2 = 2.5 t = 2500 kg ; k2 = 0.6 m ; d2 = 1.5 m or r2 = 0.75 m ; v = 9 km/h = 2.5 m/s; s = 6 m

Kinetic energy of rotation of the wheels and axles

We know that mass moment of inertia of the front roller,
I1 = m1(k1)2 = 2000 (0.4)2 = 320 kg-m2

and mass moment of inertia of the rear axle together with its wheels,
I2 = m2 (k2)2 = 2500 (0.6)2 = 900 kg -m2

Angular speed of the front roller,
ω1 = v/r1 = 2.5/0.6 = 4.16 rad/s
and angular speed of rear wheels,

ω2 = v/r2 = 2.5/0.75 = 3.3 rad/s

We know that kinetic energy of rotation of the front roller,

E1 =1/2 I1 (ω1)2 = 1/2 × 320(4.16)2 2770 N-m

and kinetic energy of rotation of the rear axle together with its wheels,

E2 =1/2 I2 (ω2)2 = 1/2 × 900(3.3)2 4900 N-m

∴ Total kinetic energy of rotation of the wheels,
E = E1 + E2 = 2770 + 4900 = 7670 N-m

We know that the kinetic energy of motion (i.e. kinetic energy of translation) of the road roller,
E3 = 1/2 mv2 = 1/2 x 1200 (2.5)2 = 37500 N-m

This energy includes the kinetic energy of translation of the wheels also, because the total mass (m) has been considered.

∴ Total kinetic energy of road roller,
E4 = Kinetic energy of translation + Kinetic energy of rotation
= E3 + E = 37 500 + 7670 = 45 170 N-m

Let F = Braking force required to bring the roller to rest, in newtons.
We know that the distance travelled by the road roller,
s = 6 m … (Given)
∴ Work done by the braking force
= F × s = 6 F N-m
This work done must be equal to the total kinetic energy of road roller to bring the roller to
rest, i.e.
6 F = 45 170 or F = 45 170/6 = 7528.3 N

4. A haulage rope winds on a drum of radius 500 mm, the free end being attached to a truck. The truck has a mass of 500 kg and is initially at rest. The drum is equivalent to a mass of 1250 kg with radius of gyration 450 mm. The rim speed of the drum is 0.75 m/s before the rope tightens. By considering the change in linear momentum of the truck and in the angular momentum of the drum, find the speed of the truck when the motion becomes steady.
a) 0.502 m/s
b) 0.602 m/s
c) 0.702 m/s
d) 0.802 m/s

Answer: a
Clarification: Given : r = 500 mm = 0.5 m ; m1 = 500 kg ; m2 = 1250 kg ; k = 450 mm = 0.45 m ; u = 0.75 m/s

We know that mass moment of inertia of drum,

I2 = m2.k2 = 1250 (0.45)2 = 253 kg-m2

Let v = Speed of the truck in m/s, and
F = Impulse in rope in N-s.

We know that the impulse is equal to the change of linear momentum of the truck. Therefore
F = m1.v = 500 v N-s
and moment of impulse = Change in angular momentum of drum
i.e. F x r = I1 (ω2 − ω1) = I2(u – v/r)
500v x 0.5 = 253(0.75 – v/0.5)
or, 250v = 380 − 506v

∴ 250 v + 506 v = 380
or v = 380/756 = 0.502 m/s

5. An electric motor drives a machine through a speed reducing gear of ratio 9:1. The motor armature, with its shaft and gear wheel, has moment of inertia 0.6 kg-m2. The rotating part of the driven machine has moment of inertia 45 kg-m2. The driven machine has resisting torque of 100 N-m and the efficiency of reduction gear is 95%. Find the power which the motor must develop to drive the machine at a uniform speed of 160 r.p.m.
a) 1764 W
b) 2764 W
c) 3764 W
d) 4764 W

Answer: a
Clarification: Given : G = 9; IA = 0.6 kg-m2 ; IB = 45 kg-m2;
TB = 100 N-m; η = 95% = 0.95;
N = 160 r.p.m. ; N1 = 0 ; N2 = 60 r.p.m. TA = 30 N-m

We know that the power which the motor must develop,

P = 2πN TB/60× η
= 2π × 160 × 100/60 x 0.95
= 1764 W

6. The flywheel of a steam engine has a radius of gyration of 1 m and mass 2500 kg. The starting torque of the steam engine is 1500 N-m and may be assumed constant. Determine : Angular acceleration of the flywheel.
a) 0.6 rad/s2
b) 0.8 rad/s2
c) 0.10 rad/s2
d) none of the mentioned

Answer: b
Clarification: Given : k = 1 m ; m = 2500 kg ; T = 1500 N-m
Angular acceleration of the flywheel
Let α = Angular acceleration of the flywheel.
We know that mass moment of inertia of the flywheel,
I=m.k2 = 2500×12 = 2500 kg-m2

We also know that torque ( T ),
1500 = I .α = 2500 × α
or α = 1500 / 2500 = 0.6 rad/s2

7. The flywheel of a steam engine has a radius of gyration of 1 m and mass 2500 kg. The starting torque of the steam engine is 1500 N-m and may be assumed constant. Determine : Kinetic energy of the flywheel after 10 seconds from the start.
a) 50 kJ
b) 60 kJ
c) 45 kJ
d) none of the mentioned

Answer: c
Clarification: Given : k = 1 m ; m = 2500 kg ; T = 1500 N-m
Angular acceleration of the flywheel
Let α = Angular acceleration of the flywheel.
We know that mass moment of inertia of the flywheel,
I=m.k2 = 2500×12 = 2500 kg-m2

We also know that torque ( T ),
1500 = I .α = 2500 × α
or α = 1500 / 2500 = 0.6 rad/s2

Kinetic energy of the flywheel after 10 seconds from start
First of all, let us find the angular speed of the flywheel (ω2) after t = 10 seconds from the start (i.e. ω1 = 0 ).
We know that ω2 = ω1 + α.t = 0 + 0.6 × 10 = 6 rad/s

∴ Kinetic energy of the flywheel,
E = 1/2 I(ω2)2
= 1/2 x 2500 x 62
= 45 000J
= 45 kJ

8. Which of the following objects have momentum?
a) An electron is orbiting the nucleus of an atom.
b) A UPS truck is stopped in front of the school building.
c) The high school building rests in the middle of town.
d) None of the mentioned

Answer: a
Clarification: Momentum can be thought of as mass in motion. An object has momentum if it has its mass in motion. It matters not whether the object is of large mass or small mass, moving with constant speed or accelerating; if the object is MOVING, then it has momentum.

9. A truck driving along a highway road has a large quantity of momentum. If it moves at the same speed but has twice as much mass, its momentum is ________________
a) zero
b) quadrupled
c) doubled
d) unchanged

Answer: c
Clarification: Momentum is directly related to the mass of the object. So for the same speed, a doubling of mass leads to a doubling of momentum.

10. A ball is dropped from the same height upon various flat surfaces. For the same collision time, impulses are smaller when the most bouncing take place.
a) True
b) False

Answer: c
Clarification: Since being dropped from the same height, the balls will be moving with the same pre-collision velocity (assuming negligible air resistance). Upon collision with the ground, the velocity will have to be reduced to zero – that is, the ball will cease moving downwards. This decrease in velocity constitutes the first portion of the velocity change. If the ball bounces, then there is an additional velocity change sending the ball back upwards opposite the original direction. Thus, for the same collision time, bouncing involves a greater velocity change, a greater momentum change, and therefore a greater impulse.