# Translation and Rotation

 Question 1
The rod PQ of length $L=\sqrt{2}$ m,and uniformly distributed mass of M=10 kg, is released from rest at the position shown in the figure. The ends slide along the frictionless faces OP and OQ. Assume acceleration due to gravity, $g=10 \:m/s^{2}$. The mass moment of inertia of the rod about its centre of mass and an axis perpendicular to the plane of the figure is $(ML^{2}/12)$. At this instant, the magnitude of angular acceleration (in radian/$s^{2}$ ) of the rod is _____
 A 15 B 10.5 C 7.5 D 4
GATE ME 2017 SET-2   Engineering Mechanics
Question 1 Explanation:
\begin{aligned} L&=\sqrt{2} m \\ M&=10 \mathrm{kg}\\ g&=10 \mathrm{m} / \mathrm{s}^{2} \\ \end{aligned}
Both the surfaces are smooth
\begin{aligned} \ddot{\theta} &=\alpha=\frac{3 g}{2 L} \sin \theta \\ \alpha &=\frac{3 \times 10}{2 \times \sqrt{2}} \sin 45^{\circ} \quad\left[\because \theta=45^{\circ}\right] \\ a &=\frac{3 \times 10}{2 \times \sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{30}{4} \\ &=7.5 \mathrm{rad} / \mathrm{s}^{2} \end{aligned}
 Question 2
An inextensible massless string goes over a frictionless pulley. Two weights of 100 N and 200 N are attached to the two ends of the string. The weights are released from rest, and start moving due to gravity. The tension in the string (in N) is __________

 A 182.56 B 133.33 C 4.256 D 2.56
GATE ME 2016 SET-3   Engineering Mechanics
Question 2 Explanation:

Tension in string
\begin{aligned} &=\left[\frac{2 m_{1} m_{2}}{m_{1}+m_{2}}\right] g \\ &=\left[\frac{2 \times \frac{100}{g} \times \frac{200}{g}}{\frac{100}{g}+\frac{200}{g}}\right] g \\ &=\left[\frac{\frac{2 \times 100 \times 200}{g^{2}}}{\frac{(100+200)}{g}}\right] \times g \\ &=\frac{2 \times 100 \times 200}{300} \\ &=\frac{400}{3}=133.33 \mathrm{N} \end{aligned}
 Question 3
Figure shows a wheel rotating about $O_{2}$. Two points A and B located along the radius of wheel have speeds of 80 m/s and 140 m/s respectively. The distance between the points A and B is 300 mm. The diameter of the wheel (in mm) is ________
 A 1400mm B 1200mm C 1600mm D 1800mm
GATE ME 2015 SET-3   Engineering Mechanics
Question 3 Explanation:
\begin{aligned} \omega_{A} &=\omega_{B} \\ \frac{V_{A}}{R-0.3} &=\frac{V_{B}}{R} \\ \frac{80}{R-0.3} &=\frac{140}{R} \\ 80 R &=140 R-42 \\ 42 &=60 R \\ R &=0.7 \mathrm{m} \\ \therefore \qquad D &=2 R=1.4 \mathrm{m}=1400 \mathrm{mm} \end{aligned}
 Question 4
For the same material and the mass, which of the following configurations of flywheel will have maximum mass moment of inertia about the axis of rotation OO' passing through the center of gravity.

 A A B B C C D D
GATE ME 2015 SET-3   Engineering Mechanics
Question 4 Explanation:
$I=m r^{2}$
Here for the rimmed wheel, the mass m will be most far away distribution from the axis 0-0.
$\therefore \;(B)$ is the correct answer.
 Question 5
The initial velocity of an object is 40 m/s. The acceleration a of the object is given by the following expression:
a=-0.1v,

where v is the instantaneous velocity of the object. The velocity of the object after 3 seconds will be _______
 A 29.63m/s B 15.26m/s C 68.25m/s D 32.35m/s
GATE ME 2015 SET-2   Engineering Mechanics
Question 5 Explanation:
\begin{aligned} u &=40 \mathrm{m} / \mathrm{s} \\ a &=-0.1 \mathrm{v} \\ V &=? \\ t &=3 \mathrm{s} \\ a &=\frac{d v}{d t}=-0.1 \mathrm{v} \\ \int_{40}^{v} \frac{d v}{v} &=\int_{0}^{3}-0.1 d t \\\;[\ln v]_{40}^{v} &=-0.1[t] \\ \ln v-\ln 40 &=-0.1[3.0]=-0.3 \\ \ln v &=\ln 40-0.3 \\ \ln v &=3.38887 \\\text{or}\qquad v &=e^{3.38887}=29.63 \mathrm{m} \mathrm{s} \end{aligned}
 Question 6
A mobile phone has a small motor with an eccentric mass used for vibrator mode. The location of the eccentric mass on motor with respect to center of gravity (CG) of the mobile and the rest of the dimensions of the mobile phone are shown. The mobile is kept on a flat horizontal surface.

Given in addition that the eccentric mass = 2 grams, eccentricity = 2.19 mm, mass of the mobile = 90 grams, g = 9.81 $m/s^{2}$ . Uniform speed of the motor in RPM for which the mobile will get just lifted off the ground at the end Q is approximately
 A 3000 B 3500 C 4000 D 4500
GATE ME 2015 SET-1   Engineering Mechanics
Question 6 Explanation:
\begin{aligned} \Sigma M_{P} &=0 \text { [moment about } P=0] \\ m g \times 6 &=m_{0}\left(2.19 \times 10^{-3}\right)\left(\omega^{2}\right) \times[9] \\ 90 \times 9.81 \times 6 &=2\left(2.19 \times 10^{-3}\right) \omega^{2} \times 9 \\ 366.5836 &=\omega=\frac{2 \pi \mathrm{N}}{60} \\ \text { or } \quad N &=3500.6158 \mathrm{rpm} \end{aligned}
 Question 7
A pinion with radius $r_{1}$ , and inertia $I_{1}$ is driving a gear with radius $r_{2}$ and inertia $I_{2}$ . Torque $\tau _{1}$ is applied on pinion. The following are free body diagrams of pinion and gear showing important forces (F1 and F2) of interaction. Which of the following relations hold true?
 A $F_{1}\neq F_{2}; \tau _{1}=I_{1}\ddot{\theta }_{1};F_{2}=I_{2}\frac{r_{1}}{{r_{2}}^{2}}\ddot{\theta }_{1}$ B $F_{1}=F_{2};\tau =[I_{1}+I_{2}(\frac{r_{1}}{r_{2}})^{2}]\ddot{\theta }_{1};F_{2}=I_{2}\frac{r_{1}}{{r_{2}}^{2}}\ddot{\theta }_{1}$ C $F_{1}=F_{2};\tau =I_{1}\ddot{\theta }_{1};F_{2}=I_{2}\frac{1}{r_{2}}\ddot{\theta }_{2}$ D $F_{1}\neq F_{2};\tau =[I_{1}+I_{2}(\frac{r_{1}}{r_{2}})^{2}]\ddot{\theta }_{1};F_{2}=I_{2}\frac{r_{1}}{r_{2}}\ddot{\theta }_{1}$
GATE ME 2015 SET-1   Engineering Mechanics
Question 7 Explanation:
$\tau_{1}-F_{1} r_{1} =I_{1} \theta_{1}\qquad \ldots(i)$
$F_{1} =F_{2}\qquad\ldots(ii)$
[Action and reaction are equal by Newton's 3rd law]
$F_{2} r_{2}=\tau_{2}=I_{2} \ddot{\theta}_{2}\ldots(iii)$
From (ii) and (iii)
$F_{2}=F_{1}=\frac{I_{2} \ddot{\theta}_{2}}{r_{2}}\qquad\ldots(iv)$
From (i) and (iv)
$\tau_{1}-\left(\frac{I_{2} \ddot{\theta}_{2}}{r_{2}}\right) r_{1}=I_{1} \ddot{\theta}_{1}\qquad\ldots(v)$
We know,
$r_{1} \omega_{1}=r_{2} \omega_{2}$
$r_{1} \dot{\theta}_{1}=r_{2} \dot{\theta}_{2}$
Differentiating with respect to time
\begin{aligned} r_{1} \ddot{\theta}_{1} &=r_{2} \ddot{\theta}_{2} \\ \ddot{\theta}_{2} &=\frac{r_{1}}{r_{2}} \ddot{\theta}_{1} \\ \tau_{1} &=I_{1} \ddot{\theta}_{1}+\frac{I_{2}}{r_{2}}\left[\frac{r_{1}}{r_{2}} \ddot{\theta}_{1}\right] r_{1} \\ \tau_{1} &=\left[I_{1}+I_{2}\left[\frac{r_{1}}{r_{2}}\right]^{2}\right] \ddot{\theta}_{1} \\ \tau_{2} &=I_{2} \ddot{\theta}_{2}=F_{2} r_{2} \\ F_{2} &=\frac{I_{2}}{r_{2}} \ddot{\theta}_{2}=\frac{I_{2}}{r_{2}}\left[\frac{r_{1}}{r_{2}} \ddot{\theta}_{1}\right] \\ F_{2} &=I_{2} \frac{r_{1}}{r_{2}^{2}} \ddot{\theta}_{1} \end{aligned}
 Question 8
A swimmer can swim 10 km in 2 hours when swimming along the flow of a river. While swimming against the flow, she takes 5 hours for the same distance. Her speed in still water (in km/h) is _____
 A 3.5km/h B 4.5km/h C 2.5km/h D 1.5km/h
GATE ME 2015 SET-1   Engineering Mechanics
Question 8 Explanation:
Let velocity of woman in still water =x
Velocity of river flow =y
(i) Swimming along the river flow
Resultant velocity of woman =x+y
$\Rightarrow \quad 2=\frac{10}{x+y}$
$\Rightarrow \quad x+y=5 \qquad \ldots(i)$
(ii) Swimming againts the river flow
Resultant velocity of woman =x-y
$\Rightarrow \quad 5=\frac{10}{x-y}$
$x-y=2 \qquad \ldots(ii)$
Adding (i) and (ii)
$2 x =7$
$x =3.5 \mathrm{km} / \mathrm{h}$
 Question 9
A uniform slender rod (8 m length and 3 kg mass) rotates in a vertical plane about a horizontal axis 1 m from its end as shown in the figure. The magnitude of the angular acceleration (in rad/$s^{2}$) of the rod at the position shown is _______
 A 2.053 B 3.125 C 3.2145 D 98.6512
GATE ME 2014 SET-4   Engineering Mechanics
Question 9 Explanation:

\begin{aligned} A P &=1 \mathrm{m} \\ A C &=4 \mathrm{m} \\ P C &=3 \mathrm{m} \\ I_{C} &=\frac{m L^{2}}{12}=\frac{3 \times 8 \times 8}{12} \\ &=16 \mathrm{kg}-\mathrm{m}^{2} \\ I_{P} &=I_{C}+m(P C)^{2} \\ &=16+3 \times 3^{2}=43 \mathrm{kg}-\mathrm{m}^{2} \\ \text { Torque } &=m_{2} g \times \frac{P B}{2}-m_{1} g \frac{A P}{2} \\ T &=m_{2} \times 9.81 \times \frac{7}{2}-m_{1} \times 9.81 \times \frac{1}{2}\\ &=34.335 \mathrm{m}_{2}-4.905 \mathrm{m}_{1} \\ m_{1} &=\frac{3}{8} \mathrm{kg}, \mathrm{m}_{2}=\frac{3 \times 7}{8}=2.625 \\ T &=34.335 \times 2.625-4.905 \times 0.375 \\ &=88.29 \mathrm{N} \mathrm{m} \\ T &=I_{P} \alpha \\ \alpha &=\frac{T}{I_{p}}=\frac{88.29}{43}=2.053 \mathrm{rad} / \mathrm{s}^{2} \end{aligned}
 Question 10
Consider a flywheel whose mass M is distributed almost equally between a heavy, ring-like rim of radius R and a concentric disk-like feature of radius R/2. Other parts of the flywheel, such as spokes, etc, have negligible mass. The best approximation for $\alpha$, if the moment of inertia of the flywheel about its axis of rotation is expressed as $\alpha MR^2$, is _______
 A 0.3574 B 0.9865 C 0.5625 D 0.1236
GATE ME 2014 SET-2   Engineering Mechanics
Question 10 Explanation:

\begin{aligned} I &=\frac{M}{2}(R)^{2}+\frac{1}{2}\left(\frac{M}{2}\right)\left(\frac{R}{2}\right)^{2} \\ &=\frac{M R^{2}}{2}+\frac{1}{2} \times \frac{M}{2} \times \frac{R^{2}}{4} \\ \Rightarrow & \frac{M R^{2}}{2}+\frac{M R^{2}}{16} \\ I&=\frac{9}{16} M R^{2}=\alpha M R^{2} \\ \alpha&=\frac{9}{16}=0.5625 \end{aligned}
There are 10 questions to complete.