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}
Question 6 |
A rigid rod (AB) of length L=\sqrt{2} m is undergoing translational as well as rotational motion in the x-y plane (see the figure). The point A has the velocity V_{1}=(\hat{i}+2\hat{j}) m/s The end B is constrained to move only along the x direction
The magnitude of the velocity V_{2} (in m/s) at the end B is __________

The magnitude of the velocity V_{2} (in m/s) at the end B is __________
3.0 | |
5.268 | |
2.369 | |
3.654 |
Question 6 Explanation:

\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}
Question 7 |
A rigid link PQ is undergoing plane motion as shown in the figure (V_{P} and V_{Q} are non-zero). V_{PQ} is the relative velocity of point Q with respect to point P.
Which one of the following is TRUE?

Which one of the following is TRUE?
V_{PQ} has components along and perpendicular to PQ | |
V_{PQ} has only one component directed from P to Q | |
V_{PQ} has only one component directed from Q to P | |
V_{PQ} has only one component perpendicular to PQ |
Question 7 Explanation:

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.
Question 8 |
A bullet spins as the shot is fired from a gun. For this purpose, two helical slots as shown in the figure are cut in the barrel. Projections A and B on the bullet engage in each of the slots.

Helical slots are such that one turn of helix is completed over a distance of 0.5 m. If velocity of bullet when it exits the barrel is 20 m/s, its spinning speed in rad/s is ________

Helical slots are such that one turn of helix is completed over a distance of 0.5 m. If velocity of bullet when it exits the barrel is 20 m/s, its spinning speed in rad/s is ________
251.3274rad/s | |
300.1547rad/s | |
450.2154rad/s | |
875.2654rad/s |
Question 8 Explanation:
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}
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}
Question 9 |
A wheel of radius r rolls without slipping on a horizontal surface shown below. If the velocity of point P is 10 m/s in the horizontal direction, the magnitude of velocity of point Q (in m/s) is ______

20m/s | |
30m/s | |
45m/s | |
35m/s |
Question 9 Explanation:
\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}
Question 10 |
A circular object of radius r rolls without slipping on a horizontal level floor with the center having velocity V. The velocity at the point of contact between the object and the floor is
zero | |
V in the direction of motion | |
V opposite to the direction of motion | |
V vertically upward from the floor |
Question 10 Explanation:

\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}
There are 10 questions to complete.