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

If the plates have velocities as shown, the magnitude of the angular velocity of the disk is
\frac{v}{r} | |
\frac{v}{2r} | |
\frac{2v}{3r} | |
\frac{3v}{2r} |
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).


6.66 | |
2.82 | |
4.95 | |
8.73 |
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


5 | |
6 | |
8 | |
10 |
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

8\angle 233^{\circ} | |
10\angle 225^{\circ} | |
10\angle 217^{\circ} | |
8\angle 217^{\circ} |
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.


12 | |
16 | |
20 | |
10 |
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}
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.