Plane Motion

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}

\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}

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}

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}

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}

\begin{aligned} \sqrt{2} \sin 45^{\circ} &=1 \mathrm{m} \\ \tan \theta &=\frac{1}{2} \\ \theta &=\tan ^{-1} 0.5=26.5651^{\circ} \\ \alpha &=90^{\circ}-\left(45^{\circ}+\theta\right)=18.4349^{\circ} \\ \beta &=90^{\circ}-\left(45^{\circ}+\alpha\right)=26.5651^{\circ} \\ \frac{V_{A}}{I_{A}} &=\frac{V_{B}}{I_{B}} \\ V_{B} &=\frac{V_{A} \cdot I_{B}}{I_{A}}=\sqrt{5} \\ \cos \beta &=\frac{1}{I_{A}} \\ I_{A} &=\frac{1}{\cos 26.5651}=1.118 \mathrm{m} \\ I_{C} &=\sqrt{\left(I_{A}\right)^{2}-1^{2}} \\ &=\sqrt{(1.118)^{2}-1}=0.5 \mathrm{m} \\ \frac{V_{A}}{I_{A}} &=\frac{V_{B}}{I_{B}} \\ V_{B} &=\frac{\sqrt{3} \times[1+0.5]}{1.118}=3 \mathrm{m} / \mathrm{s} \end{aligned}

To find relative velocity direction of V_{p} is reversed
and placed at fail of \vec{V}_{Q}

Ans. (D) Resultant V_{PQ} is perpendicular to link P Q.

One turn of helix is completed over a distance of 0.5m
So bullet spins 2 \pi radian angle in 0.5m.
As speed of bullet is 20 \mathrm{m} / \mathrm{s} , so bullet completes a distance of 20 \mathrm{m} in 1 \mathrm{sec} .
In 20 \mathrm{m} bullet spins through \frac{2 \pi}{0.5} \times 20 radian
So, angular velocity =\frac{\frac{2 \pi \times 20}{0.5}}{1}=80 \pi
=251.3274 \;\mathrm{rad} / \mathrm{s}

\begin{aligned} V_{P} &=r \omega=10 \mathrm{m} / \mathrm{s} \\ V_{Q} &=2 r \omega=2 \times 10 \\ &=20 \mathrm{m} / \mathrm{s} \end{aligned}

\begin{aligned} V_{P}= & \omega \times \overline{O P}=2 \omega r \\ V_{C}= & \omega \times \overline{O C}=\omega r \\ V_{O}= & \omega \times 0=0 \end{aligned}