Category: Machine Kinematics Objective Questions

Machine Kinematics Multiple Choice Questions on “Comparison Between Involute and Cycloidal Gears”.

1. The velocity of sliding _____________ the distance of the point of contact from the pitch point.
a) is directly proportional to
b) is inversaly proportional to
c) is equal to cosɸ multiplied by
d) does not depend upon
Answer: a
Clarification: The velocity of sliding is the velocity of one tooth relative to its mating tooth along the common tangent at the point of contact.

2. In involute gears, the pressure angle is
a) dependent on the size of teeth
b) dependent on the size of gears
c) always constant
d) always variable
Answer: c
Clarification: None

3. In full depth 14⁰ involute system, the smallest number of teeth in a pinion which meshes with rack without interference is
a) 12
b) 16
c) 25
d) 32
Answer: d
Clarification: The minimum number of teeth on the pinion in order to avoid interference for 14.5⁰ full depth involute are 32 and for 20⁰ full depth involute teeth are 18.

4. The pressure angle for involute gears depends upon the size of teeth.
a) True
b) False
Answer: b
Clarification: In a gear drive, the shape of the tooth depends upon the pressure angle.

5. The contact ratio is given by
a) Length of the path of approach/Circular pitch
b) Length of the path of recess/Circular pitch
c) Length of the arc of contact/Circular pitch
d) Length of the arc of approach/cosɸ
Answer: c
Clarification: None

6. For an involute gear, the ratio of base circle radius and pitch circle radius is equal to
a) sinɸ
b) cosɸ
c) secɸ
d) cosecɸ
Answer: b
Clarification: None

7. Which of the following statement is correct for gears?
a) The addendum is less than the dedendum
b) The pitch circle diameter is the product of module and number of teeth
c) The contact ratio means the number of pairs of teeth in contact
d) All of the mentioned
Answer: d
Clarification: None

8. In a gear having involute teeth, the normal to the involute is a tangent to the
a) pitch circle
b) base circle
c) addendum circle
d) dedendum circle
Answer: b
Clarification: Addendum circle is the circle drawn through the top of the teeth and is concentric with the pitch circle.
Dedendum circle is the circle drawn through the bottom of the teeth. It is also called root circle.
Pitch circle is an imaginary circle which by pure rolling action, would give the same motion as the actual gear.

9. The centre distance between two meshing involute gears is equal to
a) sum of base circle radii/cosɸ
b) difference of base circle radii/cosɸ
c) sum of pitch circle radii/cosɸ
d) difference of pitch circle radii/cosɸ
Answer: a
Clarification: None

10. When the tip of a tooth undercuts the root on its mating gear, it is known as interference.
a) True
b) False
Answer: a
Clarification: None

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

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.

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.