# Plane Motion

 Question 1
A circular disk of radius $r$ is confined to roll without slipping at P and Q as shown in the figure. If the plates have velocities as shown, the magnitude of the angular velocity of the disk is
 A $\frac{v}{r}$ B $\frac{v}{2r}$ C $\frac{2v}{3r}$ D $\frac{3v}{2r}$
GATE ME 2020 SET-2   Engineering Mechanics
Question 1 Explanation: For pure rolling $\begin{array}{l} v_{P}=v=(P R) \omega \quad \ldots(i)\\ v_{Q}=2 v=(Q R) \omega \quad \ldots(ii) \end{array}$
Divide by (ii) to (i),
$2=\frac{Q R}{P R} \Rightarrow Q R=2(P R)$
\begin{aligned} P R+Q R &=2 r \\ P R+2(P R) &=2 r \\ P R &=\frac{2}{3} r \\ \text{From equation(i) }\quad v &=\left(\frac{2}{3} r\right) \omega \Rightarrow \omega=\frac{3 v}{2 r} \end{aligned}
 Question 2
A force of 100 N is applied to the centre of a circular disc, of mass 10 kg and radius 1 m, resting on a floor as shown in the figure. If the disc rolls without slipping on the floor, the linear acceleration (in m/$s^{2}$ ) of the centre of the disc is ________ (correct to two decimal places). A 6.66 B 2.82 C 4.95 D 8.73
GATE ME 2018 SET-2   Engineering Mechanics
Question 2 Explanation: \begin{aligned} \mathrm{m} &=10 \mathrm{kg}, R=1 \mathrm{m} \\ I &=\frac{m R^{2}}{2}=\frac{m \times 1^{2}}{2}=\frac{m}{2} \\ 100-f_{s} &=m_{a} \\ 100-f_{s} &=10 \mathrm{a} \quad\ldots(i)\\ f_{s} \times R &=I \alpha \\ f_{s} \times 1 &=\frac{m}{2} \times \alpha \\ f_{s} &=\frac{m}{2} \times a \quad\left [\begin{aligned}a=R\alpha=1\times\alpha \\a = \alpha\end{aligned} \right] \\ f_{s} &=\frac{m a}{2}\quad\ldots(ii)\\ \text{By (i) and (ii):}\quad 100-\frac{m a}{2}&=10 a\\ 100-\frac{10 \times a}{2} &=10 a \\ 100-5 a &=10 a \\ 15 a &=100 \\ a &=\frac{100}{15}=6.666 \mathrm{m} / \mathrm{s}^{2} \end{aligned}

 Question 3
A rigid rod of length 1 m is resting at an angle $\theta = 45^{\circ}$ as shown in the figure. The end P is dragged with a velocity of U = 5 m/s to the right.At the instant shown, the magnitude of the velocity V (in m/s) of point Q as it moves along the wall without losing contact is A 5 B 6 C 8 D 10
GATE ME 2018 SET-2   Engineering Mechanics
Question 3 Explanation: Treating like the elliptical trammeds. Rod motion (P, Q) (By sitting on $I_{13}$)
$\begin{array}{c} \frac{V_{P}}{I_{13} P}=\frac{V_{Q}}{I_{13} Q} \\ \frac{5}{\left(\frac{1}{\sqrt{2}}\right)}=\frac{V_{Q}}{\left(\frac{1}{\sqrt{2}}\right)} \\ V_{Q}=5 \mathrm{m} / \mathrm{s} \end{array}$
 Question 4
In a rigid body in plane motion, the point R is accelerating with respect to point P at $10\angle 180^{\circ}$ $\frac{m}{s^{2}}$. If the instantaneous acceleration of point Q is zero, the acceleration $\left (in \, \frac{m}{s^{2}} \right )$ of point R is A $8\angle 233^{\circ}$ B $10\angle 225^{\circ}$ C $10\angle 217^{\circ}$ D $8\angle 217^{\circ}$
GATE ME 2018 SET-2   Engineering Mechanics
Question 4 Explanation: As acceleration of point Q is zero, so this rigion body PQR is hinged at Q.
$\vec{a}_{R P}=\vec{a}_{R}-\vec{a}_{P}$
is given $10 \mathrm{m} / \mathrm{s}^{2}$ an angle of $180^{\circ},$ that means only radial acceleration is hence at that instant and reference is P R
\begin{aligned} a_{R P} &=(R P) \omega^{2}=10 \\ \Rightarrow\quad 20 \omega^{2} &=10\\ \Rightarrow \quad \omega&=\frac{1}{\sqrt{2}} \end{aligned}
as $\alpha$ of whole body remains same so point R has only radial acceleration at that instant
\begin{aligned} a_{R} &=Q R\left(\omega^{2}\right) \\ &=16 \times \frac{1}{2}=8 \mathrm{m} / \mathrm{s}^{2} \end{aligned}
and will be in the horizontal backward design, but our reference is only P R. So the angle of it from reference is $(180+\theta)$
\begin{aligned} \text { from } \Delta \mathrm{PQR}\quad \tan \theta&=\frac{12}{16} \\ \Rightarrow \quad \theta&= 36.8698^{\circ} \end{aligned}
So, $\quad 180+36.8698=216.8698 ; 217^{\circ}$
So answer is $8 \angle 217^{\circ}$
 Question 5
The following figure shows the velocity-time plot for a particle travelling along a straight line. The distance covered by the particle from t=0 to t=5s is _______ m. A 12 B 16 C 20 D 10
GATE ME 2017 SET-1   Engineering Mechanics
Question 5 Explanation:
V-t curve
Distance travelled is asked from t=0 to 5 s
\begin{aligned} V &=\frac{d S}{d t} \\ d S &=V d t \\ S &=\int_{0}^{5} V d t \end{aligned}
Area of V-t graph plotted on time axis from 0 to
5s
\begin{aligned} S &=\frac{1}{2} \times 1 \times 1+1 \times 1 \times \frac{1}{2}[1+4] \times 1 \\ & +\frac{1}{2}[4+2] \times 2 \\ &=10 \mathrm{m} \end{aligned}

There are 5 questions to complete.