Newton's 2nd Law applied to Rigid Bodies

Q. A car is having weight W is moving in the direction as shown in the figure. The centre of gravity (CG) of the car is located at height h from the ground, midway between the front and rear wheels. The distance between the front and rear wheels, is l. The acceleration of the car is a, and acceleration due to gravity is g. The reactions on the front wheels $\left(R_f\right)$ and rear wheels $\left(R_r\right)$ are given by

1. $R_f=\frac{W}{2}+\frac{W}{g}\left(\frac{h}{l}\right)a;R_r=\frac{W}{2}-\frac{W}{g}\left(\frac{h}{l}\right)a$

2. $R_f=R_r=\frac{W}{2}+\frac{W}{g}\left(\frac{h}{l}\right)a$

3. $R_f=R_r=\frac{W}{2}-\frac{W}{g}\left(\frac{h}{l}\right)a$

4. $R_f=\frac{W}{2}-\frac{W}{g}\left(\frac{h}{l}\right)a;R_r=\frac{W}{2}+\frac{W}{g}\left(\frac{h}{l}\right)a$

Q. A two-step pulley system is shown in figure. What will be the acceleration $a_p$ of the falling weight P? If P=90 N, Q=135 N and $r_1=2r_2$, Neglect friction and inertia of the two-step pulley.

1. 3.48 $m/s^2$
2. 1.78 $m/s^2$
3. 3.78 $m/s^2$
4. 3.56 $m/s^2$

Q. Two blocks P and Q are connected by a string, which passes over a pulley as shown in the figure. The block P is sliding on an included surface. Ignoring the masses of the string and the pulley the tension in the string is (use gravitational acceleration g =9.81 m/s${}^2$ and neglect all friction)

1. 55.2 N
2. 62.5 N
3. 74.3 N
4. 86.2 N

Q. A uniform rod of length, I = 1 m and mass, m = 2 kg is suspended with the help of two massless strings as shown in figure below. Calculate tension T (in N) in the left string at the instant when the right string snaps ($g=10m/s^2$)

1. 5 N
2. 7 N
3. 3.5 N
4. 10 N

Q.

AB and CD are two uniform and identical bars of mass 10 kg each, as shown. The hinges at A and B are frictionless. The assembly is released from rest and motion occurs in the vertical plane. At the instant that the hinge B passes the point B', the angle between the two bars will be

1. 60 degrees

2. 37.4 degrees

3. 30 degrees

4. 45 degrees

Q. A mobile phone has a small motor with an eccentric mass used for vibrator mode. The location of the eccentric mass on the motor with respect to the 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 ${\frac{m}{s}}^2.$ Uniform speed of the motor in RPM for which the mobile will get just lifted off the ground at the end Q is approximately.

1. 3000
2. 3500
3. 4000
4. 4500

Q. 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 $\left(F_{1}and{F}_2\right)$ of interaction. Which of the following relations hold true?

1. $F_1\ne F_2;{\tau }_1=I_1\ddot{\theta };F_2=I_2\frac{r_1}{r_2^2}{\ddot{\theta }}_1$
2. $F_1=F_2;{\tau }_1=I_1[I_1+I_2{\left(\frac{r_1}{r_2}\right)}^2]\ddot{\theta };F_2=I_2\frac{r_1}{r_2^2}{\ddot{\theta }}_1$
3. $F_1=F_2;{\tau }_1=I_1{\ddot{\theta }}_1;F_2=I_2\frac{1}{r_2}{\ddot{\theta }}_2$
4. $F_1\ne F_2;{\tau }_1=I_1[I_1+I_2{\left(\frac{r_1}{r_2}\right)}^2]{\ddot{\theta }}_1;F_2=I_2\frac{1}{r_2}{\ddot{\theta }}_2$

Q.

A stone of mass m at the end of a string of length I is whirled in a vertical circle at a constant speed. The tension is the string will be maximum when the stone is

1. At the top of the circle

2. Half way down from the top

3. Quarter way down from the top

4. At the bottom of the circle

Q.

A car moving with uniform acceleration covers 450 m in a 5 second interval, and covers 700 m in the next 5 second interval. The acceleration of the car is

1. $7\frac{m}{s^2}$

2. $50\frac{m}{s^2}$

3. $25\frac{m}{s^2}$

4. $10\frac{m}{s^2}$