Theory of Machines MCQ Quiz - Objective Question with Answer for Theory of Machines - Download Free PDF

Explanation:

Rack and Pinion Gear

Definition: A rack and pinion gear system is a type of linear actuator that comprises a circular gear (the pinion) engaging a linear gear (the rack). This system converts rotational motion into linear motion and is widely used in various mechanical applications.

Working Principle: In a rack and pinion system, the pinion rotates, and its teeth engage with the teeth on the rack. As the pinion turns, it moves the rack in a straight line. This conversion of rotational motion to linear motion is precise and efficient, making the rack and pinion system an essential mechanism in many engineering applications.

Advantages:

• Provides precise control of linear motion.
• Simple design and easy to manufacture.
• High efficiency in converting rotational motion to linear motion.

Disadvantages:

• Wear and tear of gears can affect the accuracy over time.
• Requires lubrication and maintenance to ensure smooth operation.

Applications: Rack and pinion systems are commonly used in steering mechanisms of vehicles, CNC machines, and other industrial equipment where precise linear motion control is required.

Explanation:

Whitworth Quick Return Mechanism

• The Whitworth Quick Return Mechanism is a mechanical device used to convert rotational motion into reciprocating motion. This mechanism is particularly known for its ability to produce a faster return stroke compared to the forward stroke, making it highly efficient for certain applications such as in shaping and slotting machines.
• The Whitworth Quick Return Mechanism consists of a rotating crank, a slotted lever (link), and a slider. When the crank rotates, it drives the slotted lever. The slider, which is attached to the tool or working element, moves back and forth due to the motion of the slotted lever. The geometry of the mechanism ensures that the time taken for the forward stroke is longer than the time taken for the return stroke.
• By fixing link 2 and making link 3 the crank, the resulting motion and characteristics align with the quick return mechanism, where the return stroke is quicker than the forward stroke. This forms Whitworth Quick Return Mechanism.

Advantages:

• Increased efficiency due to the faster return stroke, reducing the overall cycle time of the machine.
• Simplicity in design and ease of implementation in various mechanical systems.

Disadvantages:

• Non-uniform motion of the slider, which might not be suitable for applications requiring constant speed.
• Potential for increased wear and tear due to the quick motion during the return stroke.

Applications: The Whitworth Quick Return Mechanism is commonly used in shaping machines, slotting machines, and other machinery where a faster return stroke enhances productivity.

Concept:

We analyze the kinematics of a Slider-Crank mechanism to determine the angular velocity of the connecting rod when the crank rotates at a given angular speed.

Given:

• Stroke length = \(2R\) (implies crank radius = \(R\))
• Connecting rod length = \(L\)
• Ratio \(n = \frac{L}{R}\)
• Crank angular velocity = \(\omega\)
• Crank angle = \(\theta\) (measured from inner dead center)

Step 1: Establish the geometric relationship

For a Slider-Crank mechanism, the displacement \(x\) of the piston from inner dead center is:

\(x = R(1 - \cos\theta) + L\left(1 - \sqrt{1 - \left(\frac{R}{L}\sin\theta\right)^2}\right)\)

Step 2: Differentiate to find velocity relationship

The angular velocity of the connecting rod (\(\omega_{cr}\)) is found by differentiating the angular position of the connecting rod with respect to time.

The connecting rod angle \(\phi\) relates to crank angle \(\theta\) by:

\(\sin\phi = \frac{R}{L}\sin\theta = \frac{\sin\theta}{n}\)

Step 3: Derive angular velocity expression

Differentiating both sides with respect to time:

\(\cos\phi \cdot \frac{d\phi}{dt} = \frac{\cos\theta}{n} \cdot \omega\)

Thus:

\(\omega_{cr} = \frac{d\phi}{dt} = \frac{\omega \cos\theta}{n \cos\phi}\)

Using \(\cos\phi = \sqrt{1 - \sin^2\phi} = \sqrt{1 - \left(\frac{\sin\theta}{n}\right)^2}\):

\(\omega_{cr} = \frac{\omega \cos\theta}{\sqrt{n^2 - \sin^2\theta}}\)

Concept:

For Simple Harmonic Motion (SHM), the displacement of a follower is given by:

\( s = \frac{h}{2} \left(1 - \cos\left(\frac{\pi \theta}{\beta}\right)\right) \)

Where,

• \( h = 40 \, \text{mm} \) (total rise),
• \( \beta = 90^\circ = \frac{\pi}{2} \, \text{rad} \) (angle for rise),
• \( \omega = 30 \, \text{rad/s} \) (cam angular velocity).

The velocity is the time derivative of displacement:

\( v = \frac{ds}{dt} = \frac{h}{2} \times \frac{\pi}{\beta} \times \omega \times \sin\left(\frac{\pi \theta}{\beta}\right) \)

The maximum velocity occurs when \( \sin\left(\frac{\pi \theta}{\beta}\right) = 1 \), thus:

\( v_{\text{max}} = \frac{h}{2} \times \frac{\pi}{\beta} \times \omega \)

Calculation:

Substitute the values:

\( v_{\text{max}} = \frac{40}{2} \times \frac{\pi}{\pi/2} \times 30 = 20 \times 2 \times 30 = 1200 \, \text{mm/s} \)

Convert to m/s:

\( v_{\text{max}} = \frac{1200}{1000} = \boxed{1.2 \, \text{m/s}} \)

Concept:

Power developed by a rotating shaft is given by: \( P = T_{\text{mean}} \cdot \omega \)

Where, \( T_{\text{mean}} \) is the mean torque and \( \omega \) is the angular speed in rad/s.

Given:

Torque, \(T(\theta) = 800 + 180 \sin(3\theta) \, \text{Nm}\)

Speed, N = \(600 \, \text{rpm} , \pi = 3.14\)

Calculation:

Since the average of a sine function over a complete cycle is 0, the mean torque:

\( T_{\text{mean}} = 800 \, \text{Nm} \)

Angular velocity, \( \omega = \frac{2\pi N}{60} = \frac{2 \times 3.14 \times 600}{60} = 62.8 \, \text{rad/s} \)

Power, \( P = 800 \times 62.8 = 50240 \, \text{W} = 50.24 \, \text{kW} \)

Explanation:

Mechanism and Inversion:

• When one of the links of a kinematic chain is fixed, the chain is known as a mechanism.
• A mechanism with four links is known as a simple mechanism, and a mechanism with more than four links is known as a compound mechanism.
• We can obtain as many mechanisms as the number of links in the kinematic chain by fixing, in turn, different links in a kinematic chain.
• This method of obtaining different mechanisms by fixing different links in a kinematic chain is known as the inversion of a mechanism.

Explanation:

• A kinematic chain is a group of links either joined together or arranged in a manner that permits them to move relative to one another. 

Now, let's take an example of such a mechanism

The above-shown figure is a cam and follower mechanism. 

In this mechanism, there are 2 lower pairs i.e. between links (1,2 and 3,1) and 1 higher pair which is between link (2,3). 

Hence, the minimum number of links that can make a mechanism is 3.

Mistake Points

If it is given in the question that all the pairs are lower pairs then the correct answer will be 4.

If the links are connected in such a way that no motion is possible, it results in a locked chain or structure.

A minimum number of three links is required to form a closed chain but there will be no relative motion between these three links if all are lower pairs. Hence it cannot form a kinematic chain.

So the minimum of four links is necessary to form a kinematic chain when all the lower pairs are used.

Topper’s approach:

The follower turns = 18 × 16/8 = 36 times

Detailed Solution:

Let, the follower turns = x times

According to the question,

⇒ x × 8 = 16 × 18

⇒ x = 16 × 18/8

⇒ x = 36

Concept:

Kutzback equation for DOF is given by

DOF = 3(n - 1) - 2j - h

where n = Number of links, j = Number of joints, h = Number of higher pairs.

Calculation:

Given:

From fig.

n = 10, j = 12, h = 0

DOF can be calculated as 

DOF = 3(n - 1) - 2j - h

DOF= 3(10 - 1) - (2 × 12) - 0

∴ DOF = 3

The number of tertiary links are 3 as shown below.

Explanation:

Kinematic pairs are classified under three headings namely, lower pair, higher pair and wrapping pair.

Lower Pair: 

A pair is said to be a lower pair when the connection between two elements is through the area of contact. Some of the types of Lower pair are:

• Revolute Pair
• Prismatic Pair
• Screw Pair
• Cylindrical Pair
• Planar Pair

Higher Pair: 

A pair is said to be higher pair when the connection between two elements has only a point or line of contact. Examples of higher pairs are:

• A point contact takes place when spheres rest on plane or curved surfaces (in case of ball bearings)
• Contact between teeth of a skew-helical gears
• Contact made by roller bearings
• Contact between teeth of most of the gears
• Contact between cam-follower
• Spherical Pair

Wrapping Pairs:

In a higher pair, the contact between the two bodies has only a line contact or a point contact. Whereas in a wrapping pair, one body completely wraps over the other. The typical example is of a belt and a pulley or a chain and a sprocket where the belt completely wraps around the pulley or the chain completely wraps around the sp

Spherical Pair:

This pair usually look like lower pair but it is multiple point contact pair like wrapping pair which is a higher pair.

• Example. Ball in Socket joint.

Concept:

Mechanical advantage:

• It is a number that tells us how many times a simple machine multiplies the effort force. 
• is defined as the ratio of output force to input force.
• The mechanical advantage of a machine gives its efficiency. 
• Formula, \(MA=\frac{F_0}{F_i}\), where, F₀ = output force, F_i = input force
• It is a unitless and dimensionless quantity.

Efficiency:

• The efficiency of a machine is the ratio of the work done on the load by the machine to the work done on the machine by the effort.
• Thus, it is the ratio of useful work done by the machine output to the work done by the machine input.
• It is represented by the Greek symbol η.
• Formula, efficiency, \(\eta=\frac{mechanical\, \, advantage}{velocity \, \, ratio}\)

Velocity ratio:

• The ratio of the distance moved by the point at which the effort is applied in a simple machine to the distance moved by the point at which the load is applied at the same time.
• Formula, \(velocity\, \, ratio=\frac{distance\, \, moved\,\, by\,\, effort}{distance \,\, moved\,\, by\,\, load}\)
• In the case of an ideal (frictionless and weightless) machine, velocity ratio = mechanical advantage.

Calculation:

Given: mechanical advantage, MA = 2.5, length of load lift, L_R = 2.5 m, length of effort, L_E = 10 m

Velocity ratio, \(v=\frac{10}{2.5}=4\)

Efficiency, \(\eta=\frac{mechanical \, \, advantage}{velocity \,\, ratio}\)

\(\eta=\frac{2.5}{4}=0.625\)

In percentage, the efficiency is 62.50 %.

Concept:

Mechanical Advantage: 

• In a simple machine when the effort(P) balances a load (W), the ratio of the load to the effort is called MA.
• \(M.A = \frac{{Load}}{{Effort}} = \frac{W}{P}\)

Velocity Ratio: 

• It is the ratio between the distance moved by the effort to the distance moved by the load.
• \(V.R = \frac{{distance\;moved\;by\;the\;effort\;\left( {{d_P}} \right)}}{{distance\;moved\;by\;the\;load\;\left( {{d_w}} \right)}}\)

Efficiency: 

• It is the ratio of output to input. In a simple mechanism, it is also defined as the ratio of mechanical advantage to the velocity ratio.
• \(\eta = \frac{{Output}}{{Input}} = \frac{{M.A}}{{V.R}}\)
• In actual machines, a mechanical advantage is less than the velocity ratio
• In an ideal machine, the mechanical advantage is equal to the velocity ratio.

Calculation:

Given:

Effort (P) = 15 units, MA = 3

\(M.A = \frac{W}{P}\)

\(3 = \frac{W}{15}\)

Load, W = 45 units

Explanation:

Reverted gear train:

When the axes of the first gear and the last gear are co-axial, then the gear train is known as a reverted gear train. The reverted gear trains are used in automotive transmissions, lathe back gears, industrial speed reducers, and in clocks (where the minute and hour hand shafts are co-axial).

Advantages:

• The epicyclic gear trains are useful for transmitting high-velocity ratios with gears of moderate size in a comparatively lesser space.
• The epicyclic gear trains are used in the back gear of lathe, differential gears of the automobiles, hoists, pulley blocks, wrist-watches, etc.

Explanation:

Governor Effort

• It is the force exerted by the governor at the sleeve as the sleeve tends to move.
• When the speed of the governor is constant, the force exerted on the sleeve is zero as the sleeve doesn't tend to move and hence at the constant speed, the effort of the governor is zero, but when the speed changes and the sleeve tends to move to new equilibrium position force is exerted on the sleeve.
• This force gradually diminishes to zero as the sleeves move to a new equilibrium position corresponding to the new speed.
• The mean force exerted on the sleeve during the given change of speed is known as the effort of the governor.
• The given change of speed is generally taken as 1%, hence effort is defined as the force exerted on the sleeve for 1% change of speed.

Explanation:

Isochronous Governor: 

• A governor is said to be isochronous when the equilibrium speed is constant (i.e. range of speed is zero) for all radii of rotation of the balls within the working range, neglecting friction. 
• The isochronism is the stage of infinite sensitivity.
• A spring-loaded governor can only possibly be an Isochronous governor.

Hunting: 

• If a governor is too sensitive, a small variation of speed causes large change in the sleeve movement. Thus, a governor is said to be hunt if the speed of the engine fluctuates continuously above and below the mean speed.

Stability: 

• A governor is said to be stable when for each speed within the working range there is only one radius of rotation of the fly-ball at which governor is in equilibrium.