250+ TOP MCQs on Kinematic Link or Element and Answers

Machine Kinematics Multiple Choice Questions on “Kinematic Link or Element”.

1. In a reciprocating steam engine, which of the following forms a kinematic link?
a) cylinder and piston
b) piston and connecting rod
c) crankshaft and flywheel
d) flywheel and engine frame
Answer: c
Clarification: Each part of a machine which moves relative to some other part, is known as a kinematic link. The piston and cylinder in a steam engine, form a pair and the motion of the piston is limited to a definite direction relative to the cylinder irrespective of the direction of motion of the crank.

2. A link or element need not to be a rigid body, but it must be a resistant body.
a) True
b) False
Answer: a
Clarification: A body is said to be a resistant body if it is capable of transmitting the required forces with negligible deformation. A link or element need not to be a rigid body, but it must be a resistant body.

3. A railway bridge is an example of a machine.
a) True
b) False
Answer: b
Clarification: When a mechanism is required to transmit power or to do some particular type of work, then it becomes a machine. A railway bridge remains static and there is no motion in it. Hence, it is not a machine.

4. The motion between a pair when limited to a definite direction, irrespective of the direction of force applied, is known as
a) completely constrained motion
b) incompletely constrained motion
c) successfully constrained motion
d) none of the mentioned
Answer: a
Clarification: When the motion between a pair is limited to a definite direction irrespective of the direction of force applied, then the motion is said to be a completely constrained motion.

5. The motion between a pair which takes place in ____________ is known as incompletely constrained motion.
a) one direction only
b) more than one direction
c) opposite direction
d) none of the mentioned
Answer: b
Clarification: When the motion between a pair can take place in more than one direction, then the motion is called an incompletely constrained motion.

6. When the connection between the elements forming a pair is such that the constrained motion is not completed by itself, but by some other means, the motion is said to be a completely constrained motion.
a) True
b) False
Answer: b
Clarification: When the motion between a pair is limited to a definite direction irrespective of the direction of force applied, then the motion is said to be a completely constrained motion.
When the connection between the elements forming a pair is such that the constrained motion is not completed by itself, but by some other means, the motion is said to be a successfully constrained motion.

7. The example of completely constrained motion is
a) motion of a piston in the cylinder of a steam engine
b) motion of a square bar in a square hole
c) motion of a shaft with collars at each end in a circular hole
d) all of the mentioned
Answer: d
Clarification: When the motion between a pair is limited to a definite direction irrespective of the direction of force applied, then the motion is said to be a completely constrained motion. In all the above mechanism, motion is limited, so they all are examples of completely constrained motion.

8. The motion of a shaft in a circular hole is an example of
a) completely constrained motion
b) incompletely constrained motion
c) successfully constrained motion
d) none of the mentioned
Answer: b
Clarification: When the motion between a pair can take place in more than one direction, then the motion is called an incompletely constrained motion. A circular bar or shaft in a circular hole, is an example of an incompletely constrained motion as it may either rotate or slide in a hole.

9. The example of successfully constrained motion is a
a) motion of an I.C. engine valve
b) motion of the shaft between a foot-step bearing
c) piston reciprocating inside an engine cylinder
d) all of the mentioned
Answer: d
Clarification: When the connection between the elements forming a pair is such that the constrained motion is not completed by itself, but by some other means, the motion is said to be a successfully constrained motion. so, the above examples have successfully constrained motion.

10. Which of the following statement is wrong?
a) A round bar in a round hole form a turning pair
b) A square bar in a square hole form a sliding pair
c) A vertical shaft in a foot step bearing forms a successful constraint
d) All of the mentioned
Answer: c
Clarification: None

250+ TOP MCQs on Double Slider Crank Chain & its Inversions and Answers

Machine Kinematics written test Questions & Answers on “Double Slider Crank Chain & its Inversions”.

1. Which of the following instruments is used to draw ellipses?
a) Elliptical trammels
b) Slotted lever and crank
c) Gnome engine
d) Oldham’s coupling
Answer: a
Clarification: Elliptical trammel is an instrument which is an inversion of a double slider crank chain. It is mainly used to draw ellipses.

2. Elliptical trammels are used to convert reciprocating motion into rotary motion.
a) True
b) False
Answer: b
Clarification: Elliptical trammels are used to draw ellipses; scotch yoke mechanisms are used to convert rotary motion into reciprocating motion.

3. How inversion is obtained in an elliptical trammel?
a) Fixing the slotted plate
b) Fixing the sliders
c) Fixing the turning pairs
d) Fixing the pin
Answer: a
Clarification: An elliptical trammel is an instrument which is an inversion of a double slider crank chain, here the inversion is obtained by fixing the slotted plate. It is used to draw ellipses.

4. In the given figure, 1 and 2 are sliders, 3 is a bar and 4 is a fixed slotted plate, identify the mechanism.
machine-kinematics-written-test-questions-answers-q4
a) Elliptical trammel
b) Scotch yoke mechanism
c) Oldham’s coupling
d) Gnome engine
Answer: a
Clarification: In the given figure, the slotted lever is fixed and there are two sliders, hence it is a double slider crank chain, since the slotted plate is fixed, it is an inversion. This inversion is known as elliptical trammels.

5. In the given figure if P is not the midpoint of the line connecting 1 and 2, what is the locus of P?
machine-kinematics-written-test-questions-answers-q4
a) Ellipse
b) Straight line
c) Parabola
d) Rectangular hyperbola
Answer: a
Clarification: The given figure represents an elliptical trammel, in an elliptical trammel any point on the bar traces a path which is an ellipse, hence the locus of P is ellipse.

6. In the given figure if P is the midpoint of the line connecting 1 and 2, what is the locus of P?
machine-kinematics-written-test-questions-answers-q4
a) Ellipse
b) Circle
c) Parabola
d) Rectangular hyperbola
Answer: b
Clarification: The given figure represents an elliptical trammel, in an elliptical trammel any point on the bar traces a path which is an ellipse, hence the locus of P is ellipse, but in this case P is the midpoint hence, the locus of P will be a circle.

7. Which of the mechanism is used to convert rotary motion into a reciprocating motion?
a) Elliptical trammel
b) Scotch yoke mechanism
c) Oldham’s coupling
d) Gnome engine
Answer: b
Clarification: Scotch yoke mechanism is an example of an inversion of a double slider crank chain, this mechanism is used to convert rotary motion into reciprocating motion.

8. In Scotch yoke mechanism, the crank is fixed in order to obtain the inversion.
a) True
b) False
Answer: b
Clarification: In scotch yoke mechanism there are in total 4 links, one is Crank and one is frame. Inversion can be obtained by fixing either of the remaining two links.

9. Which of the following mechanism is used for connecting two parallel shafts whose axes are at a small distance apart?
a) Elliptical trammel
b) Scotch yoke mechanism
c) Oldham’s coupling
d) Gnome engine
Answer: c
Clarification: Oldham’s coupling is an instrument which is an inversion of double slider crank chain, this is used to connect two parallel shafts whose axes are at a small distance apart.

10. Which of the following link is fixed to obtain inversion in Oldham’s coupling?
a) Driving shaft
b) Flange
c) Supporting frame
d) Driven shaft
Answer: c
Clarification: The shafts are coupled in such a way that if one shaft rotates, the other shaft also rotates at the same speed. This inversion is obtained by fixing the link which corresponds to Supporting frame.

11. How many turning pairs are there in a double slider crank chain?
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: Double slider crank chain is a kinematic chain which consists of two turning pairs, as a result it is called double slider crank chain.

12. A double slider crank chain has one pair of each sliding and turning pairs.
a) True
b) False
Answer: b
Clarification: In a double slider crank chain, there are two turning pairs and two sliding pairs. In a single slider crank chain there is one sliding pair and three turning pairs.

13. How many sliding pairs are there in a double slider crank chain?
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: Double slider crank chain is a kinematic chain which consists of two sliding pairs, as a result it is called double slider crank chain.

14. In which of the following mechanisms inversion is obtained by fixing the cylinder?
a) Pendulum pump
b) Gnome engine
c) Double slider crank chain
d) Oscillating cylinder
Answer: a
Clarification: Pendulum pump also known as the bull engine is an inversion of single slider crank chain, here inversion is obtained by fixing the cylinder.

15. Which of the following is not an inversion of double slider crank chain?
a) Elliptical trammels
b) Scotch yoke mechanism
c) Oldham’s coupling
d) Gnome engine
Answer: d
Clarification: Gnome engine is an example of inversion of single slider crank chain, whereas Elliptical trammels, Scotch yoke mechanism and Oldham’s coupling are inversions of double slider crank chain.

To practice all written questions on Machine Kinematics,

250+ TOP MCQs on Exact Straight Line Motion Consisting of One Sliding Pair and Answers

Machine Kinematics Question Paper on “Exact Straight Line Motion Consisting of One Sliding Pair”.

1. Which of the following is an exact straight line mechanism?
a) Scott Russell’s mechanism
b) Watt’s mechanism
c) Grasshopper mechanism
d) Robert’s mechanism
Answer: a
Clarification: Scott Russell’s mechanism is an exact straight line mechanism, however this mechanism is not so useful for practical purposes as friction and wear of sliding pair is more than that of turning pair.

2. Scott Russell’s mechanism is very important for practical purposes.
a) True
b) False
Answer: b
Clarification: Scott Russell’s mechanism is not so useful for practical purposes as friction and wear of sliding pair is more than that of turning pair.

3. Which of the following mechanisms is an approximate straight line mechanism?
a) Scott Russell’s mechanism
b) Watt’s mechanism
c) Gnome engine
d) Oscillating engine
Answer: b
Clarification: Watt’s mechanism is an approximate straight line mechanism. It is a crossed four bar chain mechanism used in early steam engines.

4. Which of the following mechanism forms an elliptical trammel?
a) Modified Scott Russell’s mechanism
b) Watt’s mechanism
c) Gnome engine
d) Oscillating engine
Answer: a
Clarification: Modified Scott Russell’s mechanism is similar to Scott russell’s mechanism but in this case the path traced by an arbitrary point P on the link is an ellipse.

5. Which of the following mechanisms has the form of trapezium in its mean position?
a) Scott Russell’s mechanism
b) Watt’s mechanism
c) Grasshopper mechanism
d) Robert’s mechanism
Answer: d
Clarification: Robert’s mechanism is a four bar chain mechanism, which, in its mean position, has the form of a trapezium.

6. Which of the following mechanisms was used in early days to give long stroke with a very short crank?
a) Scott Russell’s mechanism
b) Watt’s mechanism
c) Grasshopper mechanism
d) Robert’s mechanism
Answer: c
Clarification: The Grasshopper mechanism was used in early days as an engine mechanism which gave long stroke with a very short crank.

7. In Scott Russell’s mechanism the straight line motion is generated.
a) True
b) False
Answer: b
Clarification: In Scott Russell’s mechanism, straight line motion is not generated but merely copied and this mechanism is not much of practical value due to wear and friction in sliding pair.

To practice all questions papers on Machine Kinematics,

250+ TOP MCQs on Friction Clutches and Answers

Machine Kinematics Multiple Choice Questions on “Friction Clutches”.

1. A jaw clutch is essentially a
a) positive action clutch
b) cone clutch
c) friction clutch
d) disc clutch
Answer: a
Clarification: The positive clutches are used when a positive drive is required. The simplest type of a positive clutch is a jaw or claw clutch.

2. The material used for lining of friction surfaces of a clutch should have _____________ coefficient of friction.
a) low
b) high
c) medium
d) none of the mentioned
Answer: b
Clarification: The material should have a high and uniform coefficient of friction.

3. The torque developed by a disc clutch is given by
a) T = 0.25 µ.W.R
b) T = 0.5 µ.W.R
c) T = 0.75 µ.W.R
d) T = µ.W.R
Answer: d
Clarification: T = µ.W.R
where W = Axial force with which the friction surfaces are held together ;
µ = Coefficient of friction ; and
R = Mean radius of friction surfaces.

4. In case of a multiple disc clutch, if n1 are the number of discs on the driving shaft and n2 are the number of the discs on the driven shaft, then the number of pairs of contact surfaces will be
a) n1 + n2
b) n1 + n2 – 1
c) n1 + n2 + 1
d) none of the mentioned
Answer: b
Clarification: If n1 are the number of discs on the driving shaft and n2 are the number of the discs on the driven shaft, then the number of pairs of contact surfaces will be
n1 + n2 – 1.

5. The cone clutches have become obsolete because of
a) small cone angles
b) exposure to dirt and dust
c) difficulty in disengaging
d) all of the mentioned
Answer: d
Clarification: A cone clutch, was extensively used in automobiles, but now-a-days it has been replaced completely by the disc clutch. In a cone clutch, the driver is keyed to the driving shaft by a sunk key and has an inside conical surface or face which exactly fits into the outside conical surface of the driven.

6. The axial force (We) required for engaging a cone clutch is given by
a) Wn sin α
b) Wn (sin α + µ cos α )
c) Wn (sin α + 0.25 µ cos α )
d) none of the mentioned
Answer: c
Clarification: Axial force required for engaging the clutch = Wn (sin α + 0.25 µ cos α )
where Wn = Normal force acting on the contact surfaces,
α = Face angle of the cone, and
µ = Coefficient of friction.

7. In a centrifugal clutch, the force with which the shoe presses against the driven member is the ___________ of the centrifugal force and the spring force.
a) difference
b) sum
c) ratio
d) none of the mentioned
Answer: a
Clarification: The force with which the shoe presses against the driven member is the difference of the centrifugal force and the spring force. The increase of speed causes the shoe to press harder and enables more torque to be transmitted.

8. Centrifugal clutches are designed to provide automatic and smooth engagement of load to driving member.
a) True
b) False
Answer: a
Clarification: Since the operating centrifugal force is a function of square of angular velocity, the friction torque for accelerating a load is also a function of square of speed driving member.

9. For a power screw having square threads with lead angle of 450 and coefficient of friction of 0.15 between screw and nut, the efficiency of the power screw, neglecting collar friction, is given by
a) 74%
b) 64%
c) 54%
d) 44%
Answer: a
Clarification: ȵ = tanα/tan(α + ɸ)
ɸ = angle of friction,
α = Helix angle or lead angle
tanɸ = 0.15
ɸ = 8.530
ȵ = 74% .

250+ TOP MCQs on Spur Gears and Answers

Machine Kinematics Multiple Choice Questions on “Spur Gears”.

1. The gears are termed as medium velocity gears, if their peripheral velocity is
a) 1–3 m / s
b) 3–15 m / s
c) 15–30 m / s
d) 30–50 m / s
Answer: b
Clarification: The gears having velocity less than 3 m/s are termed as low velocity gears and gears having velocity between 3 and 15 m / s are known as medium velocity gears. If the velocity of gears is more than 15 m / s, then these are called high speed gears.

2. The size of gear is usually specified by
a) pressure angle
b) pitch circle diameter
c) circular pitch
d) diametral pitch
Answer: b
Clarification: It is the diameter of the pitch circle. The size of the gear is usually specified by the pitch circle diameter. It is also called as pitch diameter.

3. A spur gear with pitch circle diameter D has number of teeth T. The module m is defined as
a) m = d / T
b) m = T / D
c) m = π D / T
d) m = D.T
Answer: a
Clarification: Module is the ratio of the pitch circle diameter in millimetres to the number of teeth. It is usually denoted by m. Mathematically,
Module, m = D / T.

4. In a rack and pinion arrangement, the rack has teeth of _______________ shape.
a) square
b) trepazoidal
c) oval
d) circular
Answer: b
Clarification: None.

5. The radial distance from the _______________ to the clearance circle is called working depth.
a) addendum circle
b) dedendum circle
c) pitch circle
d) none of the mentioned
Answer: a
Clarification: Working depth is radial distance from the addendum circle to the clearance circle. It is equal to the sum of the addendum of the two meshing gears.

6. The product of the diametral pitch and circular pitch is equal to
a) 1
b) 1/π
c) π
d) π × No. of teeth
Answer: c
Clarification: Diametral pitch, pd =T/D = π/pc
Circular pitch, pc = πD/T
Therefore, pd x pc = π.

7. The backlash for spur gears depends upon
a) module
b) pitch line velocity
c) tooth profile
d) both (a) and (b)
Answer: d
Clarification: Backlash is the difference between the tooth space and the tooth thickness, as measured on the pitch circle.

8. The contact ratio for gears is
a) zero
b) less than one
c) greater than one
d) none of the mentioned
Answer: c
Clarification: The ratio of the length of arc of contact to the circular pitch is known as contact ratio i.e. number of pairs of teeth in contact.

9. If the centre distance of the mating gears having involute teeth is increased, then the pressure angle
a) increases
b) decreases
c) remains unchanged
d) none of the mentioned
Answer: a
Clarification: The pressure angle increases with the increase in centre distance.

10. The form factor of a spur gear tooth depends upon
a) circular pitch only
b) pressure angle only
c) number of teeth and circular pitch
d) number of teeth and the system of teeth
Answer: d
Clarification: Form factor is independent of the size of the tooth and depends only on the number of teeth on a gear and the system of teeth.

11. Lewis equation in spur gears is used to find the
a) tensile stress in bending
b) shear stress
c) compressive stress in bending
d) fatigue stress
Answer: c
Clarification: None.

12. The minimum number of teeth on the pinion in order to avoid interference for 20° stub system is
a) 12
b) 14
c) 18
d) 32
Answer: b
Clarification: None.

13. The allowable static stress for steel gears is approximately ____________ of the ultimate tensile stress.
a) one-fourth
b) one-third
c) one-half
d) double
Answer: b
Clarification: The allowable static stress (σo) for steel gears is approximately one-third of the ultimate tensile strength
u) i.e. σo = σu / 3.

14. Lewis equation in spur gears is applied
a) only to the pinion
b) only to the gear
c) to stronger of the pinion or gear
d) to weaker of the pinion or gear
Answer: d
Clarification: The Lewis equation is applied only to the weaker of the two wheels (i.e. pinion or gear).

15. The static tooth load should be ____________ the dynamic load.
a) less than
b) greater than
c) equal to
d) none of the mentioned
Answer: b
Clarification: For safety, against tooth breakage, the static tooth load should be greater than the dynamic load.

250+ TOP MCQs on Numericals On Kinematics Of Motion and Answers

Machine Kinematics Multiple Choice Questions on “Numericals On Kinematics Of Motion”.

1. A car starts from rest and accelerates uniformly to a speed of 72 km. p.h. over a distance of 500 m. Calculate the acceleration.
a) 0.3 m/s2
b) 0.4 m/s2
c) 0.5 m/s2
d) 0.6 m/s2

Answer: b
Clarification: Given : u = 0 ; v = 72 km. p.h. = 20 m/s ; s = 500 m
First of all, let us consider the motion of the car from rest.
Acceleration of the car
Let a = Acceleration of the car.
We know that
v2 = u2 + 2as
or, 202 = 0 + 2a x 500 = 1000a
or, a = 202/1000 = 0.4 m/s2

2. A car starts from rest and accelerates uniformly to a speed of 72 km. p.h. over a distance of 500 m. Calculate the time taken to attain the speed.
a) 50 s
b) 60 s
c) 70 s
d) 80 s

Answer: a
Clarification: Given : u = 0 ; v = 72 km. p.h. = 20 m/s ; s = 500 m
First of all, let us consider the motion of the car from rest.
Acceleration of the car
Let a = Acceleration of the car.
We know that
v2 = u2 + 2as
or, 202 = 0 + 2a x 500 = 1000a
or, a = 202/1000 = 0.4 m/s2

Let t = Time taken by the car to attain the speed.
We know that v = u + a.t
∴ 20 = 0 + 0.4 × t or t = 20/0.4 = 50 s

3. A car starts from rest and accelerates uniformly to a speed of 72 km. p.h. over a distance of 500 m. If a further acceleration raises the speed to 90 km. p.h. in 10 seconds, find this acceleration and the further distance moved.
a) 0.3 m/s2
b) 0.4 m/s2
c) 0.5 m/s2
d) 0.6 m/s2

Answer: c
Clarification: Given : u = 0 ; v = 72 km. p.h. = 20 m/s ; s = 500 m
First of all, let us consider the motion of the car from rest.
Acceleration of the car
Let a = Acceleration of the car.
We know that
v2 = u2 + 2as
or, 202 = 0 + 2a x 500 = 1000a
or, a = 202/1000 = 0.4 m/s2

Let t = Time taken by the car to attain the speed.
We know that v = u + a.t
∴ 20 = 0 + 0.4 × t or t = 20/0.4 = 50 s

Now consider the motion of the car from 72 km.p.h. to 90 km.p.h. in 10 seconds.
Given : Initial velocity, u = 72 km.p.h. = 20 m/s ;
Final velocity, v = 96 km.p.h. = 25 m/s ; t = 10 s
Let a = Acceleration of the car.
We know that v = u + a.t
25 = 20 + a × 10 or a = (25 – 20)/10 = 0.5 m2

4. A car starts from rest and accelerates uniformly to a speed of 72 km. p.h. over a distance of 500 m. A further acceleration raises the speed to 90 km. p.h. in 10 seconds.The brakes are now applied to bring the car to rest under uniform retardation in 5 seconds. Find the distance travelled during braking.
a) 200 m
b) 300 m
c) 225 m
d) 335 m

Answer: c
Clarification: Given : u = 0 ; v = 72 km. p.h. = 20 m/s ; s = 500 m
First of all, let us consider the motion of the car from rest.
Acceleration of the car
Let a = Acceleration of the car.
We know that
v2 = u2 + 2as
or, 202 = 0 + 2a x 500 = 1000a
or, a = 202/1000 = 0.4 m/s2

Let t = Time taken by the car to attain the speed.
We know that v = u + a.t
∴ 20 = 0 + 0.4 × t or t = 20/0.4 = 50 s

Now consider the motion of the car from 72 km.p.h. to 90 km.p.h. in 10 seconds.
Given : Initial velocity, u = 72 km.p.h. = 20 m/s ;
Final velocity, v = 96 km.p.h. = 25 m/s ; t = 10 s
Let a = Acceleration of the car.

We know that v = u + a.t
25 = 20 + a × 10 or a = (25 – 20)/10 = 0.5 m2

We know that distance moved by the car,
s = ut + 1/2 at2
= 20 x 10 + 1/2 0.5(10)2 = 225 m

5. A wheel accelerates uniformly from rest to 2000 r.p.m. in 20 seconds. What is its angular acceleration?
a) 10.475 rad/s2
b) 11.475 rad/s2
c) 12.475 rad/s2
d) 13.475 rad/s2

Answer: a
Clarification: Given : N0 = 0 or ω = 0 ; N = 2000 r.p.m. or ω = 2π × 2000/60 = 209.5 rad/s ; t = 20s
Angular acceleration
Let α = Angular acceleration in rad/s2.
We know that
ω = ω0 + α.t
or 209.5 = 0 + α × 20
∴ α = 209.5 / 20 = 10.475 rad/s2

6. A wheel accelerates uniformly from rest to 2000 r.p.m. in 20 seconds.How many revolutions does the wheel make in attaining the speed of 2000 r.p.m.?
a) 333.4
b) 444.4
c) 555.4
d) 666.4
a) 10.475 rad/s2
b) 11.475 rad/s2
c) 12.475 rad/s2
d) 13.475 rad/s2

Answer: a
Clarification: Given : N0 = 0 or ω = 0 ; N = 2000 r.p.m. or ω = 2π × 2000/60 = 209.5 rad/s ; t = 20s
Angular acceleration
Let α = Angular acceleration in rad/s2.
We know that
ω = ω0 + α.t
or 209.5 = 0 + α × 20
∴ α = 209.5 / 20 = 10.475 rad/s2

We know that the angular distance moved by the wheel during 2000 r.p.m. (i.e. when ω = 209.5 rad/s),
θ = (ω0 + ω)t/2 = ( 0 + 209.5)20/2 = 2095 rad
Since the angular distance moved by the wheel during one revolution is 2π radians, therefore
number of revolutions made by the wheel,
n = θ /2π = 2095/2π = 333.4

7. A horizontal bar 1.5 metres long and of small cross-section rotates about vertical axis through one end. It accelerates uniformly from 1200 r.p.m. to 1500 r.p.m. in an interval of 5 seconds. What is the linear velocity at the beginning?
a) 288.6 m/s
b) 388.6 m/s
c) 488.6 m/s
d) 188.6 m/s

Answer: d
Clarification: Given : r = 1.5 m ; N0 = 1200 r.p.m. or ω0 = 2 π × 1200/60 = 125.7 rad/s ;
N = 1500 r.p.m. or ω = 2 π × 1500/60 = 157 rad/s ; t = 5 s

Linear velocity at the beginning
We know that linear velocity at the beginning,
v0 = r . ω0 = 1.5 × 125.7 = 188.6 m/s

8. A horizontal bar 1.5 metres long and of small cross-section rotates about vertical axis through one end. It accelerates uniformly from 1200 r.p.m. to 1500 r.p.m. in an interval of 5 seconds. What is the linear velocity at the end of the interval ?
a) 235.5 m/s
b) 335.5 m/s
c) 435.5 m/s
d) 535.5 m/s

Answer: a
Clarification: Given : r = 1.5 m ; N0 = 1200 r.p.m. or ω0 = 2 π × 1200/60 = 125.7 rad/s ;
N = 1500 r.p.m. or ω = 2 π × 1500/60 = 157 rad/s ; t = 5 s

Linear velocity at the beginning
We know that linear velocity at the beginning,
v0 = r . ω0 = 1.5 × 125.7 = 188.6 m/s

Linear velocity at the end of 5 seconds
We also know that linear velocity after 5 seconds,
v5 = r . ω = 1.5 × 157 = 235.5 m/s

9. A horizontal bar 1.5 metres long and of small cross-section rotates about vertical axis through one end. It accelerates uniformly from 1200 r.p.m. to 1500 r.p.m. in an interval of 5 seconds. What is the normal component of the acceleration of the mid-point of the bar after 5 seconds after the acceleration begins ?
a) 2.7 m/s2
b) 3.7 m/s2
c) 4.7 m/s2
d) 5.7 m/s2

Answer: c
Clarification: Given : r = 1.5 m ; N0 = 1200 r.p.m. or ω0 = 2 π × 1200/60 = 125.7 rad/s ;
N = 1500 r.p.m. or ω = 2 π × 1500/60 = 157 rad/s ; t = 5 s

Linear velocity at the beginning
We know that linear velocity at the beginning,
v0 = r . ω0 = 1.5 × 125.7 = 188.6 m/s

Linear velocity at the end of 5 seconds
We also know that linear velocity after 5 seconds,
v5 = r . ω = 1.5 × 157 = 235.5 m/s

Let α = Constant angular acceleration.
We know that ω = ω0+ α.t
157 = 125.7 + α × 5 or α = (157 – 125.7) /5 = 6.26 rad/s2

Radius corresponding to the middle point,
r = 1.5 /2 = 0.75 m
∴ Tangential acceleration = α. r = 6.26 × 0.75 = 4.7 m/s2

10. A horizontal bar 1.5 metres long and of small cross-section rotates about vertical axis through one end. It accelerates uniformly from 1200 r.p.m. to 1500 r.p.m. in an interval of 5 seconds. What is the tangential component of the acceleration of the mid-point of the bar after 5 seconds after the acceleration begins ?
a) 18287 m/s2
b) 18387 m/s2
c) 18487 m/s2
d) 18587 m/s2

Answer: c
Clarification: Given : r = 1.5 m ; N0 = 1200 r.p.m. or ω0 = 2 π × 1200/60 = 125.7 rad/s ;
N = 1500 r.p.m. or ω = 2 π × 1500/60 = 157 rad/s ; t = 5 s

Linear velocity at the beginning
We know that linear velocity at the beginning,
v0 = r . ω0 = 1.5 × 125.7 = 188.6 m/s

Linear velocity at the end of 5 seconds
We also know that linear velocity after 5 seconds,
v5 = r . ω = 1.5 × 157 = 235.5 m/s

Let α = Constant angular acceleration.
We know that ω = ω0+ α.t
157 = 125.7 + α × 5 or α = (157 – 125.7) /5 = 6.26 rad/s2

Radius corresponding to the middle point,
r = 1.5 /2 = 0.75 m
∴ Tangential acceleration = α. r = 6.26 × 0.75 = 4.7 m/s2

Radial acceleration = ω2 . r = (157)2 0.75 = 18 487 m/s2